Question

The engine of a motorcycle can produce a maximum acceleration 5 m/s2. It brakes can produce a maximum retardation 10 m/s2. If motorcyclist starts from point A and reaches point B. What is the minimum time in which it can cover if the distance between A and B is 1.5 km​

Answer 1

Explanation:

Given acceleration a= 5m/s2 and maximum retardation =10 m/s2 and distance between A and B= 1.5 km= 1500 m

We know acceleration a=tv−u considering u= 0 m/s as initially the motorcycle is at rest.

So v=5t., where t is the time taken to accelerate.------(A) 

As T is the total time than time taken for retardation by an amount of 10 m/s2 be T-t. 

From equation of motion we have retardation=timeFinalvelocity−initial

In case of retardation or deceleration final velocity will be 0 as the motorcycle will come to rest.

So −v=−10(T−t)-------(B)

Putting the value of A in B we get 

−5t=−10T+10t

T=23t-------(C)

As the total distance covered by the motorcycle in T seconds=1500 m,

thus 21vT=1500

vT=3000

As earlier found v=5t and T=23t

5t23t=3000

t2=400

t=20s

we know T=23t

T=30s