Q 11-02-042JEE MainMedium
A car starts from rest at $t = 0$ with an acceleration that varies as $a = (6 - 2t)\ \text{m/s}^2$ for $0 \le t \le 3$ s, after which the acceleration becomes zero. Find the maximum speed of the car in m/s.
Numerical value type. Enter your answer.
Answer: 9
Concept. Integrate the acceleration; the speed grows while $a > 0$ and then stays constant.
Step 1. Velocity. $v = \displaystyle\int_0^t(6 - 2t)\,dt = 6t - t^2$.
Step 2. Maximum. $a \geq 0$ up to $t = 3$ s and zero afterwards:
$$v_{\max} = 6(3) - 9 = 9\ \text{m/s}$$
Answer. $9$.
Solution by Sreeraj P, M.Sc Physics