A car starts from rest at a point P at time t = 0 and moves along a straight road until it stops at a point Q. The distance x (in metres) covered by it in t seconds is \(x = t^{2}\left(3 - \displaystyle\frac{t}{2}\right)\). The time taken to reach Q and the distance PQ are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → t = 4 s, PQ = 16 m
The car stops when its velocity \(\frac{dx}{dt}\) becomes zero again. Writing \(x = 3t^{2} - \frac{t^{3}}{2}\):
\( \displaystyle v = \frac{dx}{dt} = 6t - \frac{3t^{2}}{2} \)
\( \displaystyle = \frac{3t}{2}(4 - t) \)
So v = 0 at t = 0 (the start) and at t = 4. The car reaches Q at t = 4 s, and
\( \displaystyle PQ = 4^{2}\left(3 - \frac{4}{2}\right) = 16 \times 1 = 16 \) m
Why the other options are incorrect
Option (2) t = 2 s, PQ = 8 m — Incorrect. At t = 2 the velocity is \(3(4 - 2) = 6\) m/s, so the car is still moving; t = 2 is where the acceleration \(6 - 3t\) is zero.
Option (3) t = 4 s, PQ = 32 m — Incorrect. The time is right, but x(4) = 16 × (3 − 2) = 16 m.
Option (4) t = 3 s, PQ = \(\displaystyle\frac{27}{2}\) m — Incorrect. At t = 3 the velocity is \(\frac{9}{2}(4 - 3) = \frac{9}{2}\) m/s, not zero.