Question

A car starts from rest and moves along the x-axis with constant acceleration of $5m{s}^{-2}$ for $8seconds$. If it then continues with constant velocity, what distance will the car cover in $12seconds$ since it started from the rest?

• A. $300m$
• B. $310m$
• C. $320m$
• D. $400m$

Answer 1

The correct option is C $320m$

Given,
initial velocity, $u=0$,
acceleration, $a=5m{s}^{-2}$ and
time taken, $t=8s$.

Let $s_1$ be the distance travelled and $v$ be the final velocity.

From second equation of motion, $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_1=0\times 8+\frac{1}{2}\times 5\times 8^2=160m$

From first equation of motion, $v=u+at\Rightarrow v=0+5\times 8=40m{s}^{-1}$

Now we have to find the distance covered in time $t=12s$

We know, $distance=speed\times time$.
As the speed is constant, after $8s$. i.e, speed $=40m{s}^{-1}$, therefore in the remaining 4 s its covers distance, $s_2=40\times 4=160m$
Thus total distance covered in 12 s, $s_1+s_2=160+160=320m$

Hence the correct option is (C)