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
$$v = \int_0^t (6 - 2t)\,dt = 6t - t^2$$
The acceleration is positive up to $t = 3$ s and zero afterwards, so the speed keeps increasing until $t = 3$ s and then stays constant:
$$v_{max} = 6(3) - 3^2 = 9\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics