Question

The engine of a motorcycle can produce a maximum acceleration 5m/$s^2$. Its brakes can produce a maximum retardation 10m/$s^2$. If a 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

Hint: The initial and final velocities of the motorcycle are going to be zero. In between, it will attain a velocity (v), after which brakes need to be applied, so that it stops at B. Equations of motion can be used here for simplification of the problem.

Formula used:
v=u + at
s= ut+(1/2)a$t^2$

Complete answer:
Let us break the motion of the motorcycle into two parts:
1. The motorcycle is accelerating for a time t and distance d, starting from rest attains a velocity v.
2. The motorcycle is made to retard from the velocity v to come to rest at point B. The retardation occurs over a time interval (T-t) where T is the total time taken (for motion from A to B). And the distance over which retardation occurs is (1500-d) meters.

Now, in the first part,
v= u+ at,
Where we are given that a=5 m/$s^2$ and object is at rest at A. So,
v= 5t,
Now, in the second half, we apply breaks so that motorcycle which was initially moving with v, comes to rest after a time T-t
0= v -10(T-t)
$\Rightarrow$ v= 10(T-t)
Keeping the value of v:
$\Rightarrow$ 5t = 10T - 10t
$\Rightarrow$ 15t = 10 T
$\Rightarrow$ 1.5t = T

To find the value of t, we use second equation of motion:
s= ut+(1/2)a$t^2$

For the first part of the motion, u = 0 and s = d and a = 5 m/$s^2$so,
$d = \dfrac{5t^2}{2}$

Similarly, for the second part of the motion;
$\Rightarrow$ $(1500-d) = 5t.(T-t)- \dfrac{10(T-t)^2}{2}$
Keeping the value of d and T,
$\Rightarrow$ $(1500 - \dfrac{5t^2}{2}) = 5t.(1.5t-t)- \dfrac{10(1.5t-t)^2}{2}$
Simplifying this, we get:

$\Rightarrow$ $(1500 - \dfrac{5t^2}{2}) = \dfrac{5t^2}{2}- \dfrac{5t^2}{4}$
$\Rightarrow$ $(1500 - \dfrac{5t^2}{2}) = \dfrac{5t^2}{4}$
$\Rightarrow$ $1500 = \dfrac{5t^2}{4} + \dfrac{5t^2}{2}$

$\Rightarrow$ $1500= \dfrac{15t^2}{4}$
$\Rightarrow$ $t^2 = 400$
Therefore, we get t = 20s.
But, we need to find T, so
T= 1.5t= 1.5$\times$ 20 s = 30s

So, the correct answer is “Option A”.

Note:
One should not forget to put a minus sign in the equation of motion when we are dealing with retardation. In retarding a body's motion, a force is applied opposite to the direction of the motion of the body.