Q. A car accelerates from rest at a constant rate of 2${ms}^{- 2}\ $for sometime. Then, it retards at a constant rate of 4$ms^{- 2}$and comes to rest. If the total time for which it remains in motion is 3s, what is the total distance travelled?
A) 2m
B) 3m
C) 4m
D) 6m
Explanation
(d) Using, $v = u + at$ or $v - u = at$, we find that if $\left| \overrightarrow{a} \right|\ $is, it $t$ is the time for acceleration, then $\frac{t}{2}\ $is the time for retardation
Now, $t + \frac{t}{2} = 3$ or $\frac{3t}{2} = 3$ or $t = 2s$
$S = \frac{1}{2} \times 2 \times 2 \times 2 + \frac{1}{2} \times 4 \times 1 \times 1 = (4 + 2)$m=6m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.