Question

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?

Answer 1

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