Question

If the displacement of a particle at time $t$ ($0 < t < \pi$) is given by $s = 3 \sin 2t - 6 \cos t$, then the acceleration for the values of $t$ at which its velocity is zero is:

• A. $0 \text{ units/sec}^2$
• B. $2 \text{ units/sec}^2$
• C. $3 \text{ units/sec}^2$
• D. $4 \text{ units/sec}^2$

Hint:

Solve velocity condition first, then substitute into acceleration.

Answer 1

The Correct Option is A

The displacement function is: \[ s = 3\sin 2t - 6\cos t \] Velocity is: \[ v = \frac{ds}{dt} = 6\cos 2t + 6\sin t \] For velocity zero: \[ \cos 2t + \sin t = 0 \] Using $\cos 2t = 1 - 2\sin^2 t$: \[ 1 - 2\sin^2 t + \sin t = 0 \] \[ 2\sin^2 t - \sin t - 1 = 0 \] \[ (2\sin t + 1)(\sin t - 1)=0 \] \[ \sin t = 1 \Rightarrow t = \frac{\pi}{2} \] Acceleration: \[ a = \frac{dv}{dt} = -12\sin 2t + 6\cos t \] At $t=\frac{\pi}{2}$: \[ a = -12\sin \pi + 6\cos \frac{\pi}{2} = 0 \]