Q1. Two cars A and B simultaneously start a race. Velocity of the car A varies with time according to the graph shown in the figure. It acquires a velocity 50 m/s few seconds before t = 100 s and thereafter moves with this speed. Car Bruns together with car A till both acquire a velocity 20 m/s, after this car B moves with zero acceleration for one second and then follows velocity-time profile identical to that of A with a delay of one second. In this way, car Acquires the velocity 50 m/s one second after A acquires it. How much more distance "d does the car cover in the first 100 s as compared to the car B?
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Explanation:
Velocity-time graphs of Car A and B are shown in the figure.
As far the given information, velocity of both cars are identical except Car B travels at a constant speed for 1s after attaining the speed of 20 m/s.
Time to attain velocity 20 m/s for Car B is obtained from Velocity function which is the function of time, as follows
20= 50/√65×√t, Hence time t to attain velocity 20 m/s=
10.4s.
At t = 10.4 s, velocity of car A = 20m/s
At t = 11.4 s, velocity of car A = 50/√65×√11.4,
= 20.94 m/s
Hence excess distance travelled by car A = [(20+ 20.94)/2]
= 0.47 m
Hence the answer is (a) less than 25m
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