Question

Starting from rest, Saurabh peddles his bicycle to attain a velocity of $6m{s}^{-1}$ in $30s$. Then, he applies brakes so that the velocity of the bicycle comes down to $4m{s}^{-1}$ in the next $5s$. The acceleration of the bicycle in both the cases are _____ and _____ respectively.

• A. $0.2m/s^2,1m/s^2$
• B. $-0.4m/s^2,0.2m/s^2$
• C. $1m/s^2,0.2m/s^2$
• D. $0.2m/s^2,-0.4m/s^2$

Answer 1

The correct option is D

$0.2m/s^2,-0.4m/s^2$

Case I:
Given:
Initial velocity, $u_1=0m{s}^{-1}$
Final velocity, $v_1=6m{s}^{-1}$
Time taken, $t_1=30sec$

Acceleration is defined as the ratio of change in velocity to the time taken.
$a_1=\frac{v_1-u_1}{t_1}$
$a_1=\frac{6-0}{30}$
$a_1=0.2m{s}^{-2}$

Case II :
Given:
Final velocity, $v_2=4m{s}^{-1}$
Time, $t_2=5sec$

Initial velocity $\left(u_2\right)$ of the second case is the same as final velocity $\left(v_1\right)$ of case I.
$u_2=v_1=6m/s$
Acceleration in second case is given by:
$a_2=\frac{v_2-u_2}{t_2}$
$a_2=\frac{4-6}{5}$
$a=-0.4m{s}^{-2}$