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Question

a car travels with a uniform velocity of 20m/s for 5 seconds.the brakes are applied and the car is uniformly retarted.it comes to rest in further 8seconds.calculatethe distance travelled before the brakes are applied and the distance covered after the brakes are applied.find its average speed

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Solution

Car travels at uniform velocity of 20m/s for 5 seconds before brakes are applied.
So distance travelled before brakes are applied = 20 * 5 = 100m.
Uniform retardation is applied after 5 seconds and car comes to rest in 8 seconds.
To find the disance travelled after brakes are applied. let us find the uniform retardation first.
We have v = u+at where v = final vbelocity, u = initial velocity, a = acceleration and t = time. Here we have v= 0, u = 20m/s, and t = 8 seconds.
So 0 = 20+8a=> 8a=-20=> a = -2.5 m/sec^2 (negative since it is retardation)
To find distance travelled after brakes applied we use the equation
v^2=u^2+2as where s = distance travelled
Substituting values v=0, u = 20m/s, a = -2.5m/s^2 we get
0 = (20)^2 -2(2.5)s=> 0=400-5s=> s = 80m
Hence distance travelled after brakes applied = 80m
Total distance travelled = 100+80 = 180m
Total time taken = 5+8 = 13seconds
Hence average velocity = 180/13 = 13.85 m/s.

Answer: Distance before brakes applied = 100m; distance after brakes applied = 80m; average velocity = 180/13 = 13.85 m/s

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