Question

The engine of a motorcycle can produce a maximum acceleration $5$ $m/s^2$. It brakes can produce a maximum retardation $10$ $m/s^2$. 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? (Given: that motorcycle comes to rest at B)

• A. $30$ sec
• B. $15$ sec
• C. $10$ sec
• D. $5$ sec

Answer 1

Answer: (A)

$30$ sec

Given acceleration a= 5m/$s^2$ and maximum retardation =10 m/$s^2$ and distance between A and B= 1.5 km= 1500 m

We know acceleration $a=\frac{v-u}{t}$ 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/$s^2$ be T-t.

From equation of motion we have $retardation=\frac{Finalvelocity-initial}{time}$

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

$-5t=-10T+10t$

$T=\frac{3t}{2}$-------(C)

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

thus $\frac{1}{2}vT=1500$

$vT=3000$

As earlier found v=5t and T$=\frac{3t}{2}$

$5t\frac{3t}{2}=3000$

$t^2=400$

$t=20s$

we know $T=\frac{3t}{2}$

$T=30s$