Question

A car accelerates from rest at a constant rate of $2m{s}^{-2}$ for some time. Then, it retards at a constant rate of $4m{s}^{-2}$ and comes to rest. If the total time for which it remains in motion is $3s$ , Then the total distance travelled is

• A. $2m$
• B. $3m$
• C. $4m$
• D. $6m$

Answer 1

Answer: (D)

Using $v=u+atorv-u=at,$ we find that if $\left|\overrightarrow{a}\right|$ is doubled, $t$ will be halved.
If $t$ is the time for accelerations, then $\frac{t}{2}$ is the time for retardation.
Now, $t+\frac{t}{2}=3$
or $\frac{3t}{2}=3$
$t=2s$
$S=\frac{1}{2}\times 2\times 2\times 2+\frac{1}{2}\times 4\times 1\times 1=\left(4+2\right)m=6m$