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CBSE New Pattern ~ Physics 11th (Term-I), , 29, , 03, Motion in a, Straight Line, Quick Revision, 1. Rest If the position of an object does not change, w.r.t. its surrounding with the passage of time, it is, said to be at rest. e.g. Book lying on the table, a, person sitting on a chair, etc., 2. Motion If the position of an object is continuously, changing w.r.t. its surrounding w.r.t time, then it is said, to be in the state of motion. e.g. The crawling insects,, water flowing down a dam, etc., 3. Types of Motion, On the basis of the nature of path followed, motion, is classified as, ● Rectilinear Motion The motion in which a, particle moves along a straight line is called, rectilinear motion. e.g. Motion of a sliding body, on an inclined plane., ● Circular Motion The motion in which a particle, moves in a circular path is called circular motion., e.g. A string whirled in a circular loop., ● Oscillatory Motion The motion in which a, particle moves to and fro about a given point is, known as oscillatory motion. e.g. Simple, pendulum., On the basis of the number of coordinates required, to define the motion of an object, motion is, classified as, ● One-dimensional Motion (1-D) The motion of, an object is considered as 1-D, if only one, coordinate is needed to specify the position of the, object., , Two-dimensional Motion (2-D) The, motion of an object is considered as 2-D,, if two coordinates are needed to specify, the position of the object. In 2-D motion,, the object moves in a plane. e.g. A, satellite revolving around the earth., ● Three-dimensional Motion (3-D) The, motion of an object is considered as 3-D,, if all the three coordinates are needed to, specify the position of the object., This type of motion takes place in, three-dimensional space., e.g. Butterfly flying in garden, the, motion of water molecules and motion, of kite in the sky., 4. Point Object An object is considered as, point object, if the size of the object is, much smaller than the distance travelled, by it in a reasonable duration of time., e.g. Earth can be considered as a point, object in its orbit., 5. Position It is defined as the point where, an object is situated., 6. Path Length or Distance The length of, the path covered by the object in a given, time-interval is known as its path length or, distance travelled. It is a scalar quantity,, i.e. it has only magnitude but no direction., ●

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7. Displacement The change in position of an, object in a particular direction is termed as, displacement, i.e. the difference between the, final and initial positions of the object in a, given time. It is denoted by Dx., Mathematically, it is represented by, Dx = x2 - x1, where, x 1 and x 2 are the initial and final, positions of the object, respectively., Cases, ● If x, 2 > x 1, then Dx is positive., ● If x, 1 > x 2, then Dx is negative., ● If x, 1 = x 2, then Dx is zero., It is a vector quantity as it possesses both, the, magnitude and direction., 8. Uniform Motion in a Straight Line A body, is said to be in a uniform motion, if it travels, equal distances in equal intervals of time along, a straight line. A distance (x)-time (t) graph for, uniform motion is a straight line passing, through the origin., 9. Non-uniform Motion A body is said to be in, non-uniform motion, if it travels unequal, displacements in equal intervals of time., 10. Speed The path length or the distance, covered by an object divided by the time taken, to cover that distance is called its speed., Distance travelled, Speed =, Time taken, It is a scalar quantity. The speed of the object, for a given interval of time is always positive., Unit of speed In SI (MKS) system, the unit of, speed is ms –1 and in CGS, it is cms -1., Dimensional formula [M 0 LT -1], ● Average Speed Average speed of an, object is defined as the total distance, travelled divided by the total time taken., Total distance travelled, Average speed, v av =, Total time taken, ●, , Instantaneous Speed Speed at an instant is, defined as the limit of the average speed as, the time interval ( Dt ) becomes infinitesimally, small or approaches to zero., Mathematically, instantaneous speed (v i ) at, any instant of time (t) is expressed as, , Ds, v i = Dlim, t ®0, Dt, ds, or, vi =, dt, where, ds is the distance covered in time dt., 11. Velocity The rate of change in position or, displacement of an object with time is called, the velocity of that object., Displacement, i.e. Velocity =, Time, It is a vector quantity., The velocity of an object can be positive, zero, and negative according to its displacement., Unit of velocity In CGS, the unit of velocity, is cms -1 and in MKS or SI, it is ms -1., Dimensional formula [M 0LT -1 ], ●, , ●, , Average Velocity Average velocity of a, body is defined as the change in position or, displacement ( D x ) divided by the time, interval ( Dt ) in which that displacement, occurs., Average velocity,, Total displacement ( Dx ), v av =, Total time taken ( Dt ), Instantaneous Velocity Velocity at an, instant is defined as the limit of average, velocity as the time interval ( Dt ) becomes, infinitesimally small or approaches to zero., Mathematically, instantaneous velocity (v i ) at, an instant of time (t) is given by, Dx, v i = lim, Dt ® 0 D t, dx, or, vi =, dt, where, dx is displacement for time dt., , 12. Acceleration Acceleration of a body can be, expressed as the rate of change of velocity with, time., Change in velocity, Acceleration =, Time taken, It is a vector quantity. The SI unit of, acceleration is ms –2 and in CGS system, its unit, is cm s –2. Its dimensional formula is [M 0LT –2 ] .

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13. Types of Acceleration, ● Uniform Acceleration If an object is, moving with uniform acceleration, it means, that the change in velocity is equal in equal, intervals of time., ● Non-uniform Acceleration If an object has, variable or non-uniform acceleration, it, means that, the change in velocity is unequal, in equal intervals of time., ● Average Acceleration The average, acceleration over a time interval is defined as, the change in velocity divided by the time, interval., Average acceleration,, Dv v 2 - v 1, =, a av =, Dt, t2 - t1, ●, , Instantaneous Acceleration It is defined, as the acceleration of a body at a certain, instant or the limiting value of average, acceleration when time interval becomes, very small or tends to zero. So, instantaneous, acceleration,, Dv d v, =, a inst = lim, Dt ® 0 D t, dt, , dv, is the differential coefficient of v, dt, w.r.t. t., 14. Kinematic Equations for, Uniformly Accelerated Motion If the, change in velocity of an object in each unit of, time is constant, then the object is said to be, moving with constant acceleration and such a, motion is called uniformly accelerated, motion. An object moves along a straight line, with a constant acceleration a and u be the, initial velocity at t = 0 and v be the final velocity, of the object after time (t), then, ● Velocity-Time Relation v = u + at, 1 2, ● Position-Time Relation x = ut +, at, 2, where, x is the position of the object at time t., 2, ● Position-Velocity Relation v, = u 2 + 2ax, ● Displacement of the Object in, a, nth Second s (nth ) = u + ( 2n - 1), 2, , where,, , 15. Non-uniformly Accelerated Motion, When acceleration of an object is not constant, or acceleration is a function of time, then, following relations hold for one-dimensional, motion, dx, ● v =, dt, ● dx = v dt, dv, dv, ● a =, =v, dx, dt, ● dv = a dt or vdv = adx, 16. Equations of Motion for the Motion of an, Object under Gravity, When an object is thrown upwards or fall, towards the earth under the effect of gravity, only, then its motion is called motion under, gravity., In this case, the equations of motion are given, below, v = u + (+ g ) t, Upward, motion, , g, 1, (+ g ) t2, g, 2, Downward, 2, 2, v = u + 2(+ g )h, motion, In case of upward motion, acceleration due to, gravity, g is taken as negative and for, downward motion, g is taken as positive., , h = ut +, , 17. Stopping Distance for a Vehicle When, brakes are applied to a moving vehicle, the, distance it travels before stopping is called, stopping distance., u2, Stopping distance, ds =, 2a, where, u = initial velocity of the vehicle, and a = retardation., 18. Relative Velocity in 1-D It is defined as the, time rate of change of relative position of one, object w.r.t. to another., If an object A is moving with velocity v A and an, object B is moving with velocity v B , then the, velocity of object A relative to object B is, given as v AB = v A - v B, The relative velocity of object B relative to, object A is v BA = v B - v A

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19. Different Graphs related to Motion are as, follows, Displacement-Time Graph, Condition, For a stationary body, , Graph, , Graph, , Body moving with a, constant retardation and, its initial velocity is, non-zero, , B, , Time, , Body moving with a, constant retardation with, zero initial velocity, , Time, , O, , Displacement, , O, , Velocity, , Time, , Displacement, O, , Time, , O, , Body moving with a, constant retardation, , A, , O, , Body moving with, increasing acceleration, , Body moving with a, constant acceleration, , Velocity, , Displacement, , O, , Body moving with a, constant velocity, , Condition, , Body moving with, decreasing acceleration, , Time, , Velocity, , Displacement, , O, Time, , O, , Body moving with, infinite velocity, but such, motion of a body is never, possible., , Displacement, B, , Note Slope of velocity-time graph gives average, acceleration., , Acceleration-Time Graph, Condition, , O, , A, , Time, , Time, , Body moving with a, constant acceleration, , Graph, Acceleration, , Note Slope of displacement-time graph gives average, velocity., , Velocity-Time Graph, Condition, Body moving with a, constant velocity, , Velocity, , O, , Body moving with a, constant acceleration, having zero initial, velocity, , O, , Graph, , O, Time, , Body moving with, constant decreasing, acceleration, , Velocity, , O, , Body moving with, constant increasing, acceleration, , Time, , Acceleration, , O, , Time, , Time, , Acceleration, , Time

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CBSE New Pattern ~ Physics 11th (Term-I), , 33, , Objective Questions, Multiple Choice Questions, , 5. The displacement of a car is given as, , 1. Which of the following is an example of, one-dimensional motion?, (a), (b), (c), (d), , Landing of an aircraft, Earth revolving around the sun, Motion of wheels of moving train, Train running on a straight track, , 2. The coordinates of object with respect, to a frame of reference at t = 0 s are, ( - 1, 0 , 3). If t = 5 s, its coordinates are, ( - 1, 0 , 4 ), then the object is in, (a), (b), (c), (d), , motion along Z-axis, motion along X-axis, motion along Y-axis, rest position between t = 0 s and t = 5 s, , 3. A person moves towards east for 3 m,, then towards north for 4 m and then, moves vertically up by 5 m. What is his, distance now from the starting point?, (a) 5 2 m (b) 5 m, , (c) 10 m, , (d) 20 m, , 4. For a stationary object at x = 40 m, the, position-time graph is, , - 240 m, here negative sign indicates, (a), (b), (c), (d), , direction of displacement, negative path length, position of car at that point, no significance of negative sign, , 6. Snehit starts from his home and walks, 50 m towards north, then he turns, towards east and walks 40 m and then, reaches his school after moving 20 m, towards south. Then, his displacement, from his home to school is, (a) 50 m, , (b) 110 m, , (c) 80 m, , (d) 40 m, , 7. A vehicle travels half the distance l with, speed v 1 and the other half with speed, v 2 , then its average speed is, , (NCERT Exemplar), , (a), , v1 + v2, 2, , (b), , 2v 1 + v 2, v1 + v2, , (c), , 2v 1v 2, v1 + v2, , (d), , l (v 1 + v 2 ), v 1v 2, , 8. A runner starts from O and comes back, to O following path OQRO in 1h. What, is his net displacement and average, speed?, , x (m), 40, , R, , (a), , 20, 0, , 10, , 20 30 40, , t (s), O, , x (m), , (b), , Q, , 40, , 0, , t (s), , 20, , x (m), , (c), , 1km, , (a) 0,3.57 km/h, , (b) 0,0 km/h, , (c) 0,2.57 km/h, , (d) 0,1 km/h, , 9. The sign ( + ve or - ve) of the average, , 40, , velocity depends only upon, 30, , (d) None of the above, , t (s), , (a), (b), (c), (d), , the sign of displacement, the initial position of the object, the final position of the object, None of the above

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10. Find the average velocity, when a, particle completes the circle of radius, 1m in 10 s., (b) 3.14 m/s (c) 6.28 m/s (d) zero, , 11. The displacement-time graph of two, , Displacement (x), , moving particles make angles of 30°, and 45° with the X -axis. The ratio of, their velocities is, , 14. For the x-t graph given below, the v - t, graph is shown correctly in, x(m), , (a), 30°, , 45°, , (b) 1 :2, , (c) 1 :1, , (d), , 3 :2, , (b), , given below, the average velocity, between time t = 5 s and t = 7 s is, (c), , P1, , 0, , 1, , 2, , 3, , 4, , 5, , 6 7, t (s), , (a) 8 ms-1, , (b) 8.7 ms-1, , (c) 7.8 ms-1, , (d) 13.7 ms-1, , 8, , 13. Figure shows the x-t plot of a particle in, one-dimensional motion. Two different, equal intervals of time show speed in, time intervals 1 and 2 respectively, then, , t(s), , 0, , t(s), , 0, , t(s), , 0, , t(s), , v (ms–1), , P2, , (d), , 0, , v (ms–1), , Time (t), , 12. In figure, displacement-time (x - t ) graph, , 35, 30, x (m) 27.4, 25, 20, 15, 10, 5, 0, , v (ms–1), , t(s), , 0, , (a) 1 : 3, , v1 >v2, v2 >v1, v1 =v2, Data insufficient, , v (ms–1), , (a) 2 m/s, , (a), (b), (c), (d), , 15. The speed-time graph of a particle, moving along a fixed direction is as, shown in the figure. The distance, traversed by the particle between t = 0 s, to t = 10 s is, , v (ms–1), , 12, , A, , x, , C, 2, 1, , t, , O, , (a) 20 m, , 5, , (b) 40 m, , B, 10, , t (s), , (c) 60 m (d) 80 m

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16. If an object is moving in a straight line,, then, , x (m), , (a) the directional aspect of vector can be, specified by + ve and - ve signs, (b) instantaneous speed at an instant is equal, to the magnitude of the instantaneous, velocity at that instant, (c) Both (a) and (b), (d) Neither (a) nor (b), , 17. In one dimensional motion,, instantaneous speed v satisfies, (NCERT Exemplar), 0 £ v < v 0 . Then, (a) displacement in timeT must always take, non-negative values, (b) displacement x in timeT satisfies, - v 0 T < x < v 0T, (c) acceleration is always a non-negative, number, (d) motion has no turning points, , 18. The x-t equation is given as x = 2t + 1., , t1, t (s), , (a), (b), (c), (d), , 22. A particle moves in a straight line. It, can be accelerated, (a) only, if its speed changes by keeping its, direction same, (b) only, if its direction changes by keeping its, speed same, (c) Either by changing its speed or direction, (d) None of the above, , 23. An object is moving along the path, OABO with constant speed, then, , The corresponding v-t graph is, (a), (b), (c), (d), , B, , a straight line passing through origin, a straight line not passing through origin, a parabola, None of the above, , 19. The displacement x of an object is given, as a function of time, x = 2t + 3t 2 . The, instantaneous velocity of the object at, t = 2 s is, (a) 16 ms -1, (c) 10 ms -1, , (b) 14 ms -1, (d) 12 ms -1, , from rest (at t = 0) is given by, s = 6t 2 - t 3 . The time in seconds at, which the particle will attain zero, velocity again is, (b) 4, , (c) 6, , O, , A, , (a) the acceleration of the object while moving, along to path OABO is zero, (b) the acceleration of the object along the, path OA and BO is zero, (c) there must be some acceleration along the, path AB, (d) Both (b) and (c), , 20. The displacement of a particle starting, , (a) 2, , zero, positive, Data insufficient, Cannot be determined, , (d) 8, , 21. A car moves along a straight line, according to the x-t graph given below., The instantaneous velocity of the car at, t = t 1 is, , 24. The average velocity of a body moving, with uniform acceleration travelling a, distance of 3.06 m is 0.34 ms -1 . If the, change in velocity of the body is, 0.18 ms -1 during this time, its uniform, acceleration is, (a), (b), (c), (d), , 0.01 ms-2, 0.02 ms-2, 0.03 ms-2, 0.04 ms-2

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30. The resulting a-t graph for the given v-t, , 25. The slope of the straight line, connecting the points corresponding to, (v 2 , t 2 ) and (v 1 , t 1 ) on a plot of velocity, versus time gives, (a), (b), (c), (d), , average velocity, average acceleration, instantaneous velocity, None of the above, , graph is correctly represented in, , v(m/s), , 26. The displacement x of a particle at time, , (c) 2 g, , (d) -2 g, , (a), , 27. The displacement (in metre) of a, particle moving along X -axis is given, by x = 18 t + 5t 2 . The average, acceleration during the interval t 1 = 2 s, and t 2 = 4 s is, (a) 13 ms-2, (c) 27 ms-2, , (b), , (b) 10 ms-2, (d) 37 ms-2, , 28. The relation between time and distance, is t = ax + bx , where a and b are, constants. The retardation is, 2, , (a) 2 av, (c) 2 abv 3, , (c), , a (ms–2), , (b) - b + 2 g, , 4.8, 2.4, 0, – 2.4, – 4.8, – 7.2, – 9.6, –12.0, , a (ms–2), , (a) -b, , 2 4 6 8 10 12 1416 18 20 t(s), , 4.8, 2.4, 0, – 2.4, – 4.8, , a (ms–2), , t along a straight line is given by, x = a - bt + gt 2 . The acceleration of the, particle is, , (b) 2bv, (d) 2b 2v 3, , 3, , 30, 25, 24, 20, 15, 10, 5, 0, , 2 4 6 8 10 12 14 16 18 20 22, , t (s), , 2 4 6 8 10 12 14 16 18 20 22, , t (s), , 9.6, 4.8, 2.4, , 3, , 0, , 2, , 4, , 6, , 8, , 10 12, , t (s), , Velocity (cms−1), , shown in the figure. The maximum, acceleration is, , 0, , 80, , 2 4 6 8 10 12 14 16 18 20 22, , t (s), , 60, , 31. The kinematic equations of rectilinear, , 40, 20, , 10 20 30 40 50 60 70 80, Time (s), , (a) 1 cms, , (d), , a(ms–2), , 29. The v -t graph of a moving object is, , -2, , (b) 2 cms, , -2, , (d) 6 cms-2, , (c) 3 cms, , -2, , motion for constant acceleration for a, general situation, where the position, coordinate at t = 0 is non-zero, say x 0 is, (a) v = v 0 + at, 1 2, at, 2, (c) v 2 = v 02 + 2a (x - x0 ), (d) All of the above, , (b) x = x0 + v 0 t +

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36. A particle is situated at x = 3 units at, , 32. The given acceleration-time graph, , t = 0. It starts moving from rest with a, constant acceleration of 4 ms -2 . The, position of the particle at t = 3 s is, , represents which of the following, physical situations?, a, , (a) x = + 21 units, , (b) x = + 18 units, , (c) x = -21 units, , (d) None of these, , 37. Consider the relation for relative, t, , (a) A cricket ball moving with a uniform speed, is hit with a bat for a very short time, interval., (b) A ball is falling freely from the top of a, tower., (c) A car moving with constant velocity on a, straight road., (d) A football is kicked into the air vertically, upwards., , velocities between two objects A and B,, v BA = - v AB, The above equation is valid, if, (a), (b), (c), (d), , v A and v B are average velocities, v A and v B are instantaneous velocities, v A and v B are average speed, Both (a) and (b), , 38. A person is moving with a velocity of, , 10 ms . A constant force acts for 4 s on, the object and gives it a speed of 2 ms -1, in opposite direction. The acceleration, produced is, , 10 m s -1 towards north. A car moving, with a velocity of 20 ms -1 towards south, crosses the person., The velocity of car relative to the, person is, , (a) 3 ms-2, (c) 6 ms-2, , (a) - 30 ms-1, (c) 10 ms-1, , 33. An object is moving with velocity, -1, , (b) - 3 ms-2, (d) - 6 ms-2, , 34. All the graphs below are intended to, represent the same motion. One of, them does it incorrectly. Pick it up., Distance, , Velocity, , (a), , (b), , Position, Position, , (c), , Time, , 39. A motion of a body is said to be ……, if, it moves along a straight line in any, direction., (a), (b), (c), (d), , one-dimensional, two dimensional, three-dimensional, All of the above, , 40. The numerical ratio of displacement to, , Velocity, , (d), , Time, , Time, , the distance covered by an object is, always equal to or less than …… ., (a) 1, (c) Both (a) and (b), , 35. Velocity of a body moving along a, straight line with uniform acceleration a, reduces by (3/4)th of its initial velocity, in time t 0 . The total time of motion of, the body till its velocity becomes zero is, (a), , 4, t0, 3, , (b), , 3, t0, 2, , (c), , 5, t0, 3, , (d), , (b) + 20ms-1, (d) - 10 ms-1, , 8, t0, 3, , (b) zero, (d) infinity, , 41. The time taken by a 150 m long train to, cross a bridge of length 850 m is 80 s. It, is moving with a uniform velocity of, …… km/h., (a) 45, (c) 60, , (b) 90, (d) 70

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42. The distance-time graph of …… is a, straight line., (a), (b), (c), (d), , uniform motion, non-uniform motion, uniform acceleration, None of the above, , 43. Which of the following statement is, correct?, (a) The magnitude of average velocity is the, average speed., (b) Average velocity is the displacement, divided by time interval., (c) When acceleration of particle is constant,, then motion is called as non-uniformly, accelerated motion., (d) When a particle returns to its starting point,, its displacement is non-zero., , With reference to the graph, which of, the given statement(s) is/are incorrect?, (a) The instantaneous speed during the, interval t = 5 s to t = 10 s is negative at all, time instants during the interval., (b) The velocity and the average velocity for, the interval t = 0 s to t = 5 s are equal and, positive., (c) The car changes its direction of motion at, t = 5 s., (d) The instantaneous speed and the, instantaneous velocity are positive at all, time instants during the interval t = 0 s to, t = 5 s., , 46. A graph of x versus t is shown in figure., Choose correct statement given below., x, , 44. For motion of the car between t = 18 s, , A, , and t = 20 s, which of the given, statement is correct?, , B, , E, , C, , 296, , D, , x (m) 250, , 100, , O, , 10, , 18, , 20 t(s), , (a) The car is moving in a positive direction, with a positive acceleration., (b) The car is moving in a negative direction, with a positive acceleration., (c) The car is moving in positive direction with, a negative acceleration., (d) The car is moving in negative direction with, a negative acceleration., , 45. The x-t graph for motion of a car is, , 47. Match the Column I with Column II, and select the correct option from the, codes given below, Column I, , Column II, , A., , d v / dt, , p., , Acceleration, , B., , d | v |/dt, , q., , Rate of, change of, speed, , C., , dr, dt, , r., , Velocity, , D., , d | r|, dt, , s., , Magnitude, of velocity, , given below, 10, x (m), , O, , 5, , 10, t(s), , t, , (a) The particle having some initial velocity at, t = 0., (b) At point B, the acceleration a > 0., (c) At point C, the velocity and the acceleration, vanish., (d) The speed at E exceeds that at D.

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Codes, , Column I, , A, , B, , C, , D, , (a) p, , q, , r, , s, , (b) p, , r, , s, , q, , (c) q, , p, , r, , s, , (d) s, , r, , p, , q, , Position-time, graph of two, objects with, equal, velocities., , A., , 48. Given x-t graph represents the motion, of an object. Match the Column I, (parts of graph) with Column II, (representation) and select the correct, option from the codes given below., , Position-time, graph of two, objects with, unequal, velocities but, in same, direction., , B., , A, x, C, , B, O, , C., , t, , Column I, A., , Position-time, graph of two, objects with, velocities in, opposite, direction., , Column II, , Part OA of, graph, , p., , Positive, velocity, , B., , Part AB of, graph, , q., , Object at rest, , C., , Part BC of, graph, , r., , Negative, velocity, , D., , Point A in the, graph, , s., , Change in, direction of, motion, , Codes, A, , B, , C, , D, , (a) p, , q, , r, , s, , (b) p, , r, , q, , s, , (c) q, , p, , r, , s, , (d) s, , r, , q, , p, , 49. Match the Column I (position-time, graph) with Column II (representation), and select the correct option from the, codes given below., , Column II, p., , x (m), , O, , q., , t(s), , x (m), , O, , r., , t(s), , x (m), , O, , t(s), , Codes, A, , B, , C, , A, , B, , C, , (a) p, , q, , r, , (b) q, , p, , r, , (c) p, , r, , q, , (d) q, , r, , p, , Assertion-Reasoning MCQs, For question numbers 50 to 63, two, statements are given-one labelled, Assertion (A) and the other labelled, Reason (R). Select the correct answer to, these questions from the codes (a), (b), (c), and (d) are as given below, (a) Both A and R are true and R is the, correct explanation of A., (b) Both A and R are true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false and R is also false., , 50. Assertion In real-life, in a number of, situations, the object is treated as a, point object.

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Reason An object is treated as point, object, as far as its size is much smaller, than the distance, it moves in a, reasonable duration of time., , 51. Assertion If the displacement of the, body is zero, the distance covered by it, may not be zero., Reason Displacement is a vector, quantity and distance is a scalar quantity., , 52. Assertion An object can have constant, speed but variable velocity., Reason SI unit of speed is m/s., , 53. Assertion The speed of a body can be, negative., Reason If the body is moving in the, opposite direction of positive motion,, then its speed is negative., , 54. Assertion For motion along a straight, line and in the same direction, the, magnitude of average velocity is equal, to the average speed., Reason For motion along a straight line, and in the same direction, the, magnitude of displacement is not equal, to the path length., , 55. Assertion An object may have varying, speed without having varying velocity., Reason If the velocity is zero at an, instant, the acceleration is zero at that, instant., , Reason Infinite acceleration cannot be, realised in practice., , 58. Assertion In realistic situation, the x-t,, v-t and a-t graphs will be smooth., Reason Physically acceleration and, velocity cannot change values abruptly, at an instant., , 59. Assertion A body cannot be, accelerated, when it is moving, uniformly., Reason When direction of motion of, the body changes, then body does not, have acceleration., , 60. Assertion For uniform motion, velocity, is the same as the average velocity at all, instants., Reason In uniform motion along a, straight line, the object covers equal, distances in equal intervals of time., , 61. Assertion A body is momentarily at, rest at the instant, if it reverse the, direction., Reason A body cannot have, acceleration, if its velocity is zero at a, given instant of time., , 62. Assertion In the s-t diagram as shown, in figure, the body starts moving in, positive direction but not from s = 0., s, , 56. Assertion Acceleration of a moving, particle can change its direction without, any change in direction of velocity., Reason If the direction of change in, velocity vector changes, direction of, acceleration vector does not changes., , 57. Assertion The v-t graph perpendicular, to time axis is not possible in practice., , t0, , t, , Reason At t = t 0 , velocity of body, changes its direction of motion.

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63. Assertion If acceleration of a particle, , 67. If the car goes from O to P and returns, , moving in a straight line varies, as a µ t n , then s µ t n + 2 ., , back to O, the displacement of the, journey is, , Reason If a-t graph is a straight, line,then s-t graph may be a parabola., , (a) zero, (c) 420 m, , 68. The path length of journey from O to P, , Case Based MCQs, , and back to O is, , Direction Answer the questions from, 64-68 on the following case., Motion in a Straight Line, If the position of an object is continuously, changing w.r.t. its surrounding, then it is said, to be in the state of motion. Thus, motion can, be defined as a change in position of an object, with time. It is common to everything in the, universe., In the given figure, let P, Q and R represent, the position of a car at different instants of, time., R, , O, , –160 –120 –80 –40, , 0, , Q, , 40, , P, , 80 120 160 200 240 280 320 360 400 (m), X-axis, , 64. With reference to the given figure, the, position coordinates of points P and R, are, (a) P º (+ 360, 0, 0); R º (- 120, 0, 0), (b) P º (- 360, 0, 0); R º (+ 120, 0 , 0), (c) P º (0, + 360, 0); R º (- 120, 0, 0), (d) P º (0, 0, + 360); R º (0, 0, - 120), , 65. Displacement of an object can be, (a), (b), (c), (d), , positive, negative, zero, All of the above, , 66. The displacement of a car in moving, from O to P and its displacement in, moving from P to Q are, (a), (b), (c), (d), , + 360 m and - 120 m, - 120 m and + 360 m, + 360 m and + 120 m, + 360 m and - 600 m, , (b) 720 m, (d) 340 m, , (a) 0 m, (c) 360 m, , (b) 720 m, (d) 480 m, , Direction Answer the questions from, 69-73 on the following case., Average Speed and Average Velocity, When an object is in motion, its position, changes with time. So, the quantity that, describes how fast is the position changing, w.r.t. time and in what direction is given by, average velocity., It is defined as the change in position or, displacement (Dx ) divided by the time interval, (Dt ) in which that displacement occur., However, the quantity used to describe the, rate of motion over the actual path, is average, speed. It defined as the total distance travelled, by the object divided by the total time taken., , 69. A 250 m long train is moving with a, , uniform velocity of 45 kmh - 1 . The time, taken by the train to cross a bridge of, length 750 m is, (a) 56 s, (c) 80 s, , (b) 68 s, (d) 92 s, , 70. A truck requires 3 hr to complete a, journey of 150 km. What is average, speed?, (a) 50 km/h, (c) 15 km/h, , (b) 25 km/h, (d) 10 km/h, , 71. Average speed of a car between points, A and B is 20 m/s, between B and C is, 15 m/s and between C and D is 10 m/s., What is the average speed between A, and D, if the time taken in the

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mentioned sections is 20s, 10s and 5s,, respectively?, (a) 17.14 m/s, (c) 10 m/s, , (b) 15 m/s, (d) 45 m/s, , 72. A cyclist is moving on a circular track, of radius 40 m completes half a, revolution in 40 s. Its average velocity, is, (b) 2 ms -1, (d) 8 p ms-1, , (a) zero, (c) 4 p ms-1, , 74. The displacement of a body in 8 s, , 73. In the following graph, average velocity, is geometrically represented by, 35, 30, x (m) 27.4, 25, 20, 15, 10, 5, 0, , (a), (b), (c), (d), , The motion in which the acceleration remains, constant is known as to be uniformly, accelerated motion. There are certain, equations which are used to relate the, displacement (x), time taken (t ), initial velocity, (u ), final velocity (v ) and acceleration (a ) for, such a motion and are known as kinematics, equations for uniformly accelerated motion., starting from rest with an acceleration, of 20 cms -2 is, (a) 64 m, (c) 64 cm, , (b) 640 m, (d) 0.064 m, , 75. A particle starts with a velocity of, , 2 ms -1 and moves in a straight line with, a retardation of 01, . ms -2 . The first time, at which the particle is 15 m from the, starting point is, , P2, , P1, , 0, , 1, , 2, , 3, , 4, , 5, , 6 7, t (s), , 8, , length of the line P1 P2, slope of the straight line P1 P2, slope of the tangent to the curve at P1, slope of the tangent to the curve at P2, , Direction Answer the questions from, 74-78 on the following case., Uniformly Accelerated Motion, The velocity of an object, in general, changes, during its course of motion. Initially, at the, time of Galileo, it was thought that, this, change could be described by the rate of, change of velocity with distance. But, through, his studies of motion of freely falling objects, and motion of objects on an inclined plane,, Galileo concluded that, the rate of change of, velocity with time is a constant of motion for, all objects in free fall., This led to the concept of acceleration as the, rate of change of velocity with time., , (a) 10 s, (c) 30 s, , (b) 20 s, (d) 40 s, , 76. If a body starts from rest and travels, 120 cm in 6th second, then what is its, acceleration?, (a) 0.20 ms- 2, (c) 0218, . ms- 2, , (b) 0027, ., ms- 2, (d) 003, . ms- 2, , 77. An object starts from rest and moves, with uniform acceleration a. The final, velocity of the particle in terms of the, distance x covered by it is given as, (a), (c), , 2ax, ax, 2, , (b) 2ax, (d), , ax, , 78. A body travelling with uniform, acceleration crosses two points A and B, with velocities 20 ms -1 and 30 ms -1 ,, respectively. The speed of the body at, mid-point of A and B is, (a) 25 ms-1, (c) 24 ms-1, , (b) 25.5 ms-1, (d) 10 6 ms-1

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CBSE New Pattern ~ Physics 11th (Term-I), , 43, , ANSWERS, Multiple Choice Questions, 1. (d), 11. (a), , 2. (a), 12. (b), , 3. (a), 13. (b), , 4. (a), 14. (a), , 5. (a), 15. (c), , 6. (a), 16. (c), , 7. (c), 17. (b), , 8. (a), 18. (b), , 9. (a), 19. (b), , 10. (d), 20. (b), , 21. (a), 31. (d), , 22. (c), 32. (a), , 23. (d), 33. (b), , 24. (b), 34. (b), , 25. (b), 35. (a), , 26. (c), 36. (a), , 27. (b), 37. (d), , 28. (a), 38. (a), , 29. (d), 39. (a), , 30. (a), 40. (a), , 41. (a), , 42. (a), , 43. (b), , 44. (a), , 45. (a), , 46. (c), , 47. (a), , 48. (b), , 49. (b), , 52. (b), 62. (c), , 53. (d), 63. (b), , 54. (c), , 55. (d), , 56. (d), , 57. (a), , 58. (a), , 59. (d), , 66. (a), 76. (c), , 67. (a), 77. (a), , 68. (b), 78. (b), , 69. (c), , 70. (a), , 71. (a), , 72. (b), , 73. (b), , Assertion-Reasoning MCQs, 50. (a), 60. (b), , 51. (b), 61. (c), , Case Based MCQs, 64. (a), 74. (c), , 65. (d), 75. (a), , SOLUTIONS, 1. In one-dimensional motion, only one, coordinate is required to specify the position, of the object. So, a train running on a straight, track is an example of one-dimensional, motion., , 2. Given, at t = 0 s, position of an object is, ( -1 , 0 , 3) and at t = 5 s, its coordinate is, ( -1 , 0 , 4 ). So, there is no change in x and, y-coordinates, while z -coordinate changes, from 3 to 4. So,the object is in motion along, Z-axis., , 3. Distance from starting point, = ( 3) 2 + ( 4 ) 2 + ( 5) 2 = 5 2 m, , 4. For a stationary object, the position-time, graph is a straight line parallel to the time, axis, so for the given object at x = 40 m,, x-t graph is correctly shown in option (a)., , 5. In I-D motion, positive and negative signs are, used to specify the direction of motion., Since, displacement is a vector quantity, so, negative sign in -240 m indicates the, direction of displacement., , 6. Let O be the starting point, i.e. home. So,, according to the question, Snehit moves from, O to A (50 m) towards north, then from A to, , B (40 m) towards east and from B to C (20 m), towards south as shown in the figure below., 40 m, , A, , 40 m, , D, 50 m, , N, , B, 20 m, C, , E, , W, S, , 30 m, O, , θ, , P, , Displacement of Snehit is OC, which can be, calculated by Pythagoras theorem, i.e., In DODC, OC 2 = OD 2 + CD 2 = ( 30 ) 2 + ( 40 ) 2, = 900 + 1600 = 2500, Þ, , OC = 50 m, , 7. Time taken to travel first half distance,, t1 =, , l/ 2, l, =, 2v 1, v1, , Time taken to travel second half distance,, l, t2 =, 2v 2, Total time = t 1 + t 2 =, , l, l, lé1, 1ù, +, = ê + ú, 2v 1 2v 2 2 ë v 1 v 2 û

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We know that, v av = average speed, total distance, =, total time, l, 2v 1v 2, =, =, lé1, 1 ù v1 + v2, ê + ú, 2 ëv1 v2 û, , 8. As runner starts from O and comes back, to O, so net displacement is zero., Average speed, Total distance OQ + QR + RO, =, =, Total time, Total time, æ 90° ö, 1 km + ( 2pr ) ç, ÷ km + 1 km, è 360° ø, =, 1h, (Q angle of sector OQR is 90°), æ1ö, 1 + 2p ´ 1 ç ÷ + 1, è 4ø, =, 1, p, = 2 + = 3.57 km/h, 2, , 9. Since, average velocity,, v =, , D x Displacement, =, D t Time interval, , So, average velocity depends on the, displacement and hence it depends on the, sign of the displacement., , 10. When a particle completes one revolution in, circular motion, then average displacement, travelled by particle is zero., Hence, average velocity, average displacement, 0, =, =, =0, Dt, Dt, , 11. In case x - t graph is a straight line, the slope, of this line gives velocity of the particle., As slope = tan q, where q is the angle which, the tangent to the curve makes with the, horizontal in anti-clockwise direction., The velocities of two particles A and B are, 1, v A = tan 30° =, 3, v B = tan 45° = 1, The ratio of velocities,, 1, vA :vB =, :1 = 1 : 3, 3, , 12. Given, x 2 = 27.4 m, x 1 = 10 m, t 2 = 7 s and, t 1 = 5 s., Average velocity between 5 s and 7s,, x - x 1 27.4 -10, v = 2, =, t2 - t1, 7-5, =, , 17.4, = 8.7 ms -1, 2, , 13. Slope of x-t graph in a small interval, = Average speed in that interval, As, slope for interval 2 > slope for interval 1., \, v2 > v1, , 14. The x - t graph shown, is parallel to time axis., This means that, the object is at rest. So, the, velocity of the object is zero for all time, instants. Hence, v -t graph coincides with the, time axis as shown in graph (a)., , 15. Distance travelled by the particle between time, interval t = 0 s to t = 10 s, = Area of triangle OAB, 1, = ´ Base ´ Height, 2, 1, = ´ OB ´ AC, 2, 1, = ´ 10 ´ 12 = 60 m, 2, , 16. In one-dimensional motion, i.e. motion along, a straight line, there are only two directions, in which an object can move and these two, directions can be easily specified by + ve and, - ve signs., Also, in this motion instantaneous speed or, simply speed at an instant is equal to the, magnitude of instantaneous velocity at the, given instant., , 17. For maximum and minimum displacements,, we have to keep in mind the magnitude and, direction of maximum velocity., As, maximum velocity in positive direction is, v 0 and maximum velocity in opposite, direction is also - v 0 ., Maximum displacement in one direction = v 0T, Maximum displacement in opposite, directions = - v 0T, Hence, the range of displacement will be, -v 0T < x < v 0T .

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18. v =, , dx, = 2 ms -1 = constant, dt, , Since, the direction of velocity is changing,, i.e. there must be some acceleration along the, path AB., Distance, 3.06, 24. Time =, =, =9 s, Average velocity 0.34, , v ms–1, 2, t (s), , Hence, option (b) is correct., , 19. Given, x = 2t + 3t 2, dx, = 2+ 6t, dt, For t = 2 s, v = 2 + 6 ( 2) = 14 ms -1, v =, , 20. Displacement of the particle,, s = 6t 2 - t 3, Velocity of the particle,, ds d, v =, = ( 6t 2 - t 3 ), dt dt, v = 12t - 3t 2, For, v = 0 Þ 12t = 3 t 2 Þ t = 4 s, , 21. The instantaneous velocity is the slope of the, tangent to the x -t graph at that instant of, time., , x (m), , P, , t = t1, , Tangent at point P, corresponding to, t = t1, , t(s), , At t = t 1, the tangent is parallel to time axis as, shown above and hence its slope is zero., Thus, instantaneous velocity at t = t 1 is zero., , 22. Since velocity is a vector quantity, having, both magnitude and direction. So, a change, in velocity may involve change in either or, both of these factors. Therefore, acceleration, may result from a change in speed, (magnitude), a change in direction or changes, in both., , 23. For paths OA and BO, the magnitude of, velocity (speed) and direction is constant,, hence acceleration is zero. For path AB, since, this path is a curve, so the direction of the, velocity changes at every moment but the, magnitude of velocity (speed) remains, constant., , Acceleration, Change in velocity, =, Time, 0.18, =, = 0.02 ms -2, 9, , 25. Average acceleration is defined as the, average change of velocity per unit time. On, a plot of v -t, the average acceleration is the, slope of the straight line connecting the, points corresponding to ( v 2, t 2 ) and ( v 1, t 1 )., , 26. Given, x = a - bt + g t 2, dx d, = ( a - bt + g t 2 ) = - b + 2 g t, dt dt, dv d, a =, = ( - b + 2g t ) = 2g, dt dt, v =, , 27. Given, x = 18t + 5t 2, dx d, = (18t + 5t 2 ) = 18 + 10t, dt dt, \ v = 10t + 18, At t 1 = 2 s, v 1 = 10 ( 2) + 18 = 38 m/s, At t 2 = 4 s, v 2 = 10 ( 4 ) + 18 = 58 m/s, 58 - 38 20, v -v, \ a = 2 1 =, =, = 10 ms -2, 2, 2, t, 28. Given, t = ax 2 + bx, dt, = 2a x + b, dx, dx, 1, Þ, =v =, dt, 2ax + b, v =, , dv dv dx, =, dt dx dt, æ - v 2a ö, dv, 1, =, Þa =v ×, ç, ÷, dx 2ax + b è 2ax + b ø, , As, acceleration, a =, , = - 2a v × v 2 = - 2av 3, \ Retardation = 2av 3, 29. Maximum acceleration means maximum, change in velocity in minimum time interval., In time interval t = 30 s to t = 40 s,, Dv 80 - 20 60, a =, =, =, = 6 cms -2, Dt 40 - 30 10

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30. Average acceleration for different time, intervals is the slope of v-t graph, which are, as follows, ( 24 - 0 ) ms -1, For 0 s -10 s, a =, = 2.4 ms -2, (10 - 0 ) s, ( 24 - 24 ) ms -1, For 10 s -18 s, a =, = 0 ms -2, (18 - 10 ) s, ( 0 - 24 ) ms -1, For 18 s - 20 s, a =, = - 12 ms -2, ( 20 - 18 ) s, So, the corresponding a -t graph for the given, v -t graph is shown correctly in graph (a)., , 31. All the equations given in options (a), (b) and, (c) are the kinematic equations of rectilinear, motion for constant acceleration., , 32. The acceleration-time graph represents the, motion of a uniformly moving cricket ball, turned back by hitting it with a bat for a very, short time interval., , 33. Given, v = - 2 ms -1 (opposite direction),, t = 4 s and u = 10 ms -1, \ v = u + at or - 2 = 10 + 4a or a = - 3 ms -2, , 34. If velocity versus time graph is a straight line, with negative slope, then acceleration is, constant and negative., With a negative slope, distance-time graph, 1, æ, ö, will be parabolic çs = ut - at 2 ÷ ., è, 2 ø, Hence, options (a), (c) and (d) are correct, so, option (b) will be incorrect., , 35. According to kinematic equation of motion,, v = u - at, 3u, u, where, v = u = u /4 Þ, = u - at 0, 4, 4, Negatve sign signifies that the body will, decelerate, since the final velocity is, decreasing., u 4, or, = t0, a 3, u 4, Now, 0 = u - at or, t = = t0, a 3, , 36. Given, x 0 = 3 units, a = 4 ms -2, t = 3 s, Using relation, x = x 0 + v 0 t +, =3+, , 1 2, at, 2, , 1, ´ 4 ´ ( 3) 2 = + 21 units, 2, , 37. Given,, , v BA = - v AB, , The above relation is true for both average, velocities of particles and instantaneous, velocities of particles., As speed is scalar quantity, ignorant of, direction, so average speed may not be equal., , 38. Let south to north direction be positive., Velocity of car, vC = - 20 ms -1, Velocity of person, v P = + 10 ms -1, vCP = vC - v P = ( - 20 ) - (10 ) = - 30 ms -1, , 39. One-dimensional motion is a motion along, a straight line in any direction. e.g. A train is, moving on a platform., Hence, option (a) is correct., , 40. Since, displacement d is always less than or, equal to the distance D but never greater, than it, i.e. d £ D. So, numerical ratio of, displacement to the distance covered by an, object is always equal to or less than one., , 41. Total distance = Length of train + Length of, bridge, = (150 + 850 ) m = 1000 m, Distance, 1000, Time =, Þ 80 =, v, Velocity, v =, , 1000, 1000 18, m/s Þ v =, ´, = 45 km/h, 80, 80, 5, , 42. In uniform motion, the velocity of an object, does not change or it remains constant with, time., So, the graph of distance-time is a straight, line., 43. Statement given in option (b) is correct but, the rest are incorrect and these can be, corrected as,, In general, average speed is not equal to, magnitude of average velocity. It can be so, if, the motion is along a straight line without, change in direction., When acceleration of particle is not constant,, then motion is called as non-uniformly, accelerated motion., Displacement is zero, when a particle returns, to its starting point.

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44. For negative acceleration, the x - t graph, moves downward. But the car is moving in, positive direction as the position coordinate is, increasing in the positive direction., Thus, the statement given in option (a) is, correct, rest are incorrect., , 45. The instantaneous speed is always positive as, it is the magnitude of the velocity at an, instant, so it is positive during t = 5 s, to t = 10 s., For t = 0 s to t = 5 s, the motion is uniform, and x-t graph has positive slope. So, the, velocity and average velocity, instantaneous, velocity and instantaneous speed are equal, and positive., During t = 0 s to t = 5 s, the slope of the, graph is positive, hence the average velocity, and the velocity both are positive., During t = 5 s to t = 10 s, the slope of the, graph is negative, hence the velocity is, negative. Since, there is a change in sign of, velocity at t = 5 s, so the car changes its, direction at this instant., Hence, option (a) is incorrect, while all others, are correct., , dr, is the magnitude of rate of change of, dt, position of particle. This means it represents, magnitude of velocity., Hence, A ® p, B ® q, C ® r and D ® s., 48. In x -t graph, OA ® Positive slope ® Positive, velocity, AB ® Negative slope ® Negative velocity, BC ® Zero slope ® Object is at rest, At point A, there is a change in sign of, velocity, hence the direction of motion must, have changed at A., Hence, A ® p, B ® r, C ® q and D ® s., , 49., , A. For equal velocities, the slope of the, straight lines must be same as shown below, x, , θ, θ (Equal) same slope, , O, , t(s), , B. For unequal velocity, slope is different, but, since, the objects are moving in the same, direction, the slope for both the graphs, must be of same sign (positive or negative), and they meet at a point as shown below, , 46. As, point A is the starting point, therefore, particle is starting from rest., , x, θ2, , At point B, the graph is parallel to time axis,, so the velocity is constant here. Thus,, acceleration is zero., Also point C, the graph changes slope, hence, velocity also changes., After graph at C is almost parallel to time, axis, hence we can say that velocity and, acceleration vanishes., From the graph, it is clear that, slope at D > slope at E, Hence, speed at D will be more than at E ., dv, 47., is the rate of change of velocity, so it, dt, represents acceleration., d | v|, is rate of change of speed of the particle., dt, dr, is the rate by which distance of particle, dt, from the origin is changing., , θ1, , C. For velocities in opposite direction, slopes, must be of opposite sign. Slope = tan q,, where q is the angle of the straight line with, horizontal in anti-clockwise direction. As,, we know, tan q1 > 0, tan q2 < 0., Hence, slopes are of opposite sign., This condition is shown below, , x, , O, , θ1, , θ2, t(s), , Hence A ® q, B ® p and C ® r., , 50. The approximation of an object as point, object is valid only, when the size of the

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object is much smaller than the distance it, moves in a reasonable duration of time., Therefore, both A and R are true and R is, the correct explanation of A., , 51. Distance is the total path length travelled by, the object. But displacement the shortest, distance between the initial and final, positions of the object. So, distance can never, be negative or zero. But displacement can be, zero, positive and negative., Also, distance is a scalar quantity. It means, that, it is always positive but however, displacement is a vector quantity. So, it may, be positive, zero or negative depending on, given situation., Therefore, both A and R are true but R is, not the correct explanation of A., , 52. Velocity is a vector quantity, so it has both, direction and magnitude. Hence, an object, can have variable velocity by keeping its, magnitude constant, i.e. speed and by, changing direction only., The SI unit of speed is m/s., Therefore, both A and R are true but R is, not the correct explanation of A., , 53. Speed can never be negative because it is a, scalar quantity. So, if a body is moving in, negative direction, then also the speed will be, positive., , dv, = Slope of v-t graph, dt, It v-t graph is perpendicular to t-axis, slope, =¥, , 57. Acceleration, a =, , \, , a =¥, , Therefore, both A and R are true and R is, the correct explanation of A., , 58. In realistic situation, the x - t , v - t and a - t, graphs will be smooth, as the values of, acceleration and velocity cannot change, abruptly since changes are always continuous., Therefore, both A and R are true and R is, the correct explanation of A., , 59. The uniform motion of a body means that,, the body is moving with constant velocity., But if the direction of motion is changing, (such as in uniform circular motion), its, velocity changes and thus uniform, acceleration is produced in the body., Therefore, A is false and R is also false., , 60. In uniform motion along a straight line, the, object covers equal distances in equal, intervals of time., For uniform motion, x - t graph is represented, as a straight line inclined to time axis. The, average velocity during any time interval, t = t 1 to t = t 2 is the slope of the line PQ, which coincides with the graph., , Therefore, A is false and R is also false., , Q, , 54. For motion in a straight line and in the same, direction,, Displacement = Total path length, Þ Average velocity = Average speed, Therefore, A is true but R is false., , 55. If speed varies, then velocity will definitely, vary., When a particle is thrown upwards, at, highest point a ¹ 0 but v = 0., Therefore, A is false and R is also false., v - v i dv, 56. Accleration, a = f, = , i.e. direction of, Dt, dt, acceleration is same as that of change in, velocity vector or in the direction of Dv., Therefore, A is false and R is also false., , P, x, , t1, , t2, , t, , Also, velocity at any instant say t = t 1 is the, slope of the tangent at point P which again, coincides with PQ or with the graph. Hence,, velocity is same as the average velocity at all, instants., Therefore, both A and R are true but R is, not the correct explanation of A., , 61. When a particle is released from rest position, under gravity, then v = 0 but a ¹ 0.

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Also, a body is momentarily at rest at the, instant, if it reverse the direction., Therefore, A is true but R is false., , 71. Total distance ( d = vt ), , 62. Slope of s-t graph = velocity = positive, At t = 0, s ¹ 0, further at t = t 0 : s = 0, v ¹ 0., Therefore, A is true but R is false., , 63. By differentiating a-t equation two times, we, will get s-t equation., Further, a, , = 20 ´ 20 + 15 ´ 10 + 10 ´ 5 = 600 m, Total time = 20 + 10 + 5 = 35 s, Therefore, average speed, = 600 / 35 = 1714, . m/s, , 72. Given, R = 40 m and t = 40 s, Displacement, Time taken, 2R 2 ´ 40, =, =, = 2 ms -1, t, 40, , Average velocity =, , s, , 73. From the position-time graph, average, t, Straight line, , t, Parabola, , Therefore, both A and R are true but R is, not the correct explanation of A., , 64. The position coordinates of point, P = ( +360, 0, 0 ) and point R = ( -120, 0, 0 )., , 65. Displacement is a vector quantity, it can be, positive, negative and zero., , 66. Displacement, Dx = x 2 - x 1, For journey of car in moving from O to P ,, x 2 = + 360 m, x1 = 0, Þ, Dx = x 2 - x 1 = 360 - 0 = + 360 m, For journey, of car in moving from P to Q ,, x 2 = + 240 m, x 1 = + 360 m, Þ, Dx = x 2 - x 1 = 240 - 360 = - 120 m, Here, -ve sign implies that the displacement, is in –ve direction, i.e. towards left., , 67. Displacement, D x = x 2 - x 1 = 0 - 0 = 0, 68. Path length of the journey, = OP + PO = + 360 m + ( + 360 ) m = 720 m, Total distance, 69. Total time taken =, Speed, 250 + 750, = 80 s, 5, 45 ´, 18, Total distance, 70. Average speed =, Total time, 150, =, = 50 km/h, 3, t=, , velocity is geometrically represented by the, slope of curve, i.e. slope of straight line P1 P2 ., 1, 74. Displacement, s = ´ (0.2) (64) = 64 cm, 2, 1, 75. From equation of motion, s = ut - at 2, 2, 1, 15 = 2t - ´ ( 0.1)t 2 Þ t = 10 s, 2, , 76. From equation of motion,, a, ( 2n - 1), 2, a, 1.2 = 0 + ( 2 ´ 6 - 1), 2, 1.2 ´ 2, a =, = 0.218 ms - 2, 11, sn = u +, , Þ, Þ, , 77. Given,, , v0 = 0, , Using relation, v 2 = v 02 + 2ax, v 2 = 2ax, \, , v = 2ax, , 78. Let the acceleration of the car = a, and distance between A and B = d, Given, v = 30 ms -1 and u = 20 ms -1, 2ad = ( 30 ) 2 - ( 20 ) 2, 900 - 400, ad =, = 250, 2, When the car is at the mid-point of AB, then, speed of car is v 1 ., v 12 - ( 20 ) 2 = 2a ( d / 2), v 12 = ad + 400, = 250 + 400 = 650, Therefore, v 1 = 25.5 ms -1