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CASE STUDY QUESTION 03, Read the following and answer any four questions from (i) to (v), Aditya started driving his car. He increases the speed till 4 seconds and then he, kept his card in constant speed for 6 seconds. Then after he decreased the speed of, the car upto another 6 seconds. After reaching at the starting place, he draws the, speed-time graph of his 16 seconds driving as shown below:

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(i) What type of motion is represented by OA ?, (a) uniform velocity (b) uniform acceleration, (c) negative acceleration (d) no acceleration, OA is a straight line graph between speed and time, and it is, sloping upwards from O to A., Therefore, the graph line OA represents uniform, acceleration., (ii) What type of motion is represented by BC ?, (a) uniform velocity (b) uniform acceleration, (c) negative acceleration (d) no acceleration, BC is a straight line graph between speed and time which is sloping, downwards from B to C., Therefore, BC represents uniform retardation (or negative acceleration).

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(iii) Find out the acceleration of the body., (a) 1.5 m/s2, (b) 2 m/s2, (c) 3 m/s2, (d) 1 m/s2, The slope of speed-time graph OA will give us the, acceleration of the body., Thus, Acceleration = Slope of line OA, = AD/OD, Now, in the given graph, we find that AD = 6 m/s and OD, = 4 seconds., So, putting these values in the above relation, we get :, Acceleration = 6 m/s / 4 s, = 1.5 m/s2

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(iv) Calculate the retardation of the body., (a) 1.5 m/s2, (b) 2 m/s2, (c) 3 m/s2, (d) 1 m/s2, The slope of speed-time graph BC will be equal to the, retardation of the body., So, Retardation = Slope of line BC, = BE/EC, Now, in the graph given to us, we find that BE = 6, m/s and EC = 16 – 10 = 6 seconds., So, putting these values in the above relation, we get :, Retardation = 6m/s / 6 s, = 1 m/s2

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(v) Find out the distance travelled by the body from A, to B., (a) 15 m, (b) 30 m, (c) 36 m, (d) 60 m, Distance travelled from A to B, = Area under the line AB and the time axis, = Area of rectangle DABE, = DA × DE, Now, from the given graph, we find that DA = 6 m/s, and DE = 10 – 4 = 6 s., Therefore, Distance travelled from A to B = 6 × 6, = 36 m

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