Question

A body moves for a total of nine seconds starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration, and then stops. The duration of uniform acceleration.

• A. $3s$
• B. $4.5s$
• C. $5s$
• D. $6s$

Answer 1

The correct option is D

$6s$

Step 1. Given data:

Total time the body moves,$t=9s$

Uniform acceleration is $a_1$

Uniform retardation is twice acceleration,$a_2=-2a_1$

Step 2. Solving for a time the body accelerates:

$t_1$, time the body accelerates

$t_2$, time the body retards

$t=t_1+t_2=9s$

Let $V_{max}$ is the maximum velocity before the body starts to retards,

$t_1=\frac{v_{max}}{a_1}$ and $t_2=\frac{v_{max}}{2a_1}$

Now,

$t_1+t_2=\frac{v_{max}}{a_1}\left(1+\frac{1}{2}\right)=9$

$\frac{v_{max}}{a_1}\left(\frac{3}{2}\right)=9$

$v_{max}=9\times \frac{2}{3}{a}_1=6a_1$

$t_1=\frac{v_{max}}{a_1}=\frac{6a_1}{a_1}$

$t_1=6s$

Hence, the correct option is D