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www.ncrtsolutions.in, , NCERT Solutions for Class 7 Science Chapter 13, Motion and Time Class 7, Chapter 13 Motion and Time Exercise Solutions, << Previous Chapter 12 : Reproduction in PlantsNext Chapter 14 : Electric Current and its Effects >>, , Exercise : Solutions of Questions on Page Number : 156, Q1 :, Classify the following as motion along a straight line, circular oroscillatory motion:, (i) Motion of your hands while running., (ii) Motion of a horse pulling a cart on a straight road., (iii) Motion of a child in a merry-go-round., (iv) Motion of a child on a see-saw., (v) Motion of the hammer of an electric bell., (vi) Motion of a train on a straight bridge., Answer :, (i) Oscillatory motion, While running, the hands move to and fro and repeat their motion after a given interval of time. Hence, it is an oscillatory motion., (ii)Straight line, The horse is pulling a cart on a straight road. Therefore, it has a motion along a straight line., (iii)Circular motion, Merry-go-round has a circular motion. Therefore, a child sitting inside it will also have a circular motion., (iv) Oscillatory motion, The child on a see-saw goes up and down continuously. It oscillates up-down. Therefore, it is an oscillatory motion., (v)Oscillatory motion, The hammer hits the electric bell and vibrates rapidly. Therefore, it is an oscillatory motion., (vi) Straight line, The train is moving on a straight bridge. Therefore, it has a motion along a straight line., , Q2 :, Which of the following are not correct?, (i) The basic unit of time is second., (ii) Every object moves with a constant speed., (iii) Distances between two cities are measured in kilometres., (iv) The time period of a given pendulum is not constant., (v) The speed of a train is expressed in m/h., Answer :, (i) Correct, Second is the SI unit of time., (ii) Not correct, An object can move with constant or variable speed., , www.ncrtsolutions.in
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www.ncrtsolutions.in, (iii) Correct, The distance between two cities can be very large. Since kilometre is a bigger unit of distance, the distance between two cities is, measured in kilometres. (iv) Not correct, Time period of a pendulum depends on the length of the thread. Hence, it is constant for a particular pendulum., (v) Not correct, The speed of a train is measured either in km/h or in m/s., , Q3 :, A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?, Answer :, Numberof oscillations = 20, Total time taken to complete 20 oscillations = 32 s, , Q4 :, The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train., Answer :, Distance between the two stations = 240 km, Time taken = 4 h, , Q5 :, The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car, if, at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time., Express the speed in km/h also., , Answer :, Initial reading of the odometer of the car = 57321.0 km, Final reading of the odometer of the car = 57336.0 km, Distance covered by the car, = Final reading of the odometer of the car - Initial reading of the odometer of the car, = 57336.0 - 57321.0 = 15 km, The given car starts at 8:30 a.m. and stops at 8:50 a.m., Therefore, time taken by the car to cover the distance is (8:50 - 8:30) min = 20 min, Distance covered by the car = 15 km, Time taken by the car = 20 min, , www.ncrtsolutions.in
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www.ncrtsolutions.in, Again,, 60 min = 1 h, , Time taken by the car =, , Q6 :, Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the, distance between her house and the school., , Answer :, Time taken by Salma to reach her school from her home = 15 min = 15 × 60 = 900 s, Speed of her bicycle = 2 m/s, , Distance covered = Speed ×Time taken = 2 ×900 = 1800 m 1000, m = 1 km, , Q7 :, Show the shape of the distance-time graph for the motion in the following cases:, (i) A car moving with a constant speed., (ii) A car parked on a side road., Answer :, (i) A car moving with a constant speed covers equal distance in equal intervals of time. Such motion of car is represented in the, given distance-time graph., , www.ncrtsolutions.in
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www.ncrtsolutions.in, (ii)The distance-time graph of a car parked on a road side is such that with the increase in time, there is no change in distance, as, shown in the given figure., , Q8 :, Which of the following relations is correct?, (i) Speed = Distance × Time, , (ii), , (iii), , (iv), Answer :, (ii) Speed of an object is given by the relation, , Q9 :, The basic unit of speed is:, (i) km/min, (ii) m/min, (iii) km/h, (iv) m/s, Answer :, (iv)m/s, The basic unit of distance is metre (m)., The basic unit of time is second (s)., , Therefore, the basic unit of speed is m/s., , www.ncrtsolutions.in
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www.ncrtsolutions.in, Q10 :, A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total, distance covered by the car is:, (i) 100 km, (ii) 25 km, (iii) 15 km, (iv) 10 km, Answer :, (ii)25 km, Case I, , Speed of the car = 40 km/h, , Time taken = 15 min =, , Distance covered, d1= Speed ×Time taken = 40 ×0.25 = 10 km, Case II, Speed of the car = 60 km/h, , Time taken = 15 min =, , Distance covered, d2= Speed ×Time taken = 60 ×0.25 = 15 km, Total distance covered by the car, d = d1+ d2= 10 + 15 = 25 km Therefore,, the total distance covered by the car is 25 km., , Q11 :, Suppose the two photographs, shown in Figure 1 and Figure 2, had been taken at an interval of 10 seconds. If a distance of, 100 metres is shown by 1 cm in these photographs, calculate the speed of the blue car., , www.ncrtsolutions.in
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www.ncrtsolutions.in, , Figure 1 Vehicles moving in the same direction of on a road, , Figure 2 Position of vehicles shown in Figure 1 after some time, Answer :, Thedistance covered by the blue car (as evident from the photograph) from one white strip to another, which is measured by scale is, 1.4 cm., It is given that 1 cm is equivalent to 100 m., Therefore, 1.4 cm is equivalent to 140 m., Distance travelled by the car = 140 m, Time interval between the two photographs = 10 s, , Q12 :, Figure shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?, , www.ncrtsolutions.in
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www.ncrtsolutions.in, Distance-time graph for the motion of two cars, Answer :, Vehicle A is moving faster than vehicle B., , Speed is given by the relation, , This relation shows that speed of a vehicle is greater if it covers maximum distance in a given interval of time. To compare the, distance, draw a line perpendicular to the time-axis, as shown in the following distance-time graph., , From the graph, it is evident that for a given time t, the distance covered by vehicle A is more than vehicle B. Hence, vehicle A is, moving faster than vehicle B., , Q13 :, Which of the following distance-time graphs shows a truck moving with speed which is not constant?, , www.ncrtsolutions.in
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www.ncrtsolutions.in, , Answer :, Graph (iii), In a distance-time graph, the constant speed of a truck will be represented by a straight line., In a distance-time graph, a straight line parallel to the time axis indicates that the truck is not moving., A curved line on the distance-time graph indicates that the truck is moving with a speed which is not constant., , www.ncrtsolutions.in
