Question

A particle moves along a straight line such that its displacement at any time t is given by $s = (t^3-6t^2+3t+4)$ metres. The velocity when the acceleration is zero is

• A. 3 m/s
• B. 42 m/s
• C. -9 m/s
• D. -15 m/s

Answer 1

Answer: (C)

-9 m/s

Answer 2

Displacement (s) $t^3-6t^2+3t+4$ metres.
velocity (v) $=\frac{ds}{dt} = 3t^2 - 12t + 3$
acceleration (a) $=\frac{dv}{dt} = 6t - 12$
When a = 0, we get t = 2 seconds.
Therefore velocity when the acceleration is zero
$(v) \, \, = 3 \times (2)^2 -(12 \times 2) + 3 = - 9\, m/s.$