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

Step 1: Set up the motion.
The displacement is $s=3\sin2t-6\cos t$. We are asked for the acceleration at the instant when the velocity becomes zero.

Step 2: Find the velocity.
Velocity is the first derivative of displacement: \[ v=\frac{ds}{dt}=6\cos2t+6\sin t. \]
Step 3: Make the velocity zero.
Set $v=0$, so $\cos2t+\sin t=0$. Replace $\cos2t$ with $1-2\sin^2t$ to get one variable only: \[ 1-2\sin^2t+\sin t=0. \]
Step 4: Solve the quadratic in $\sin t$.
Rearrange to $2\sin^2t-\sin t-1=0$, which factors as $(2\sin t+1)(\sin t-1)=0$. So $\sin t=1$ or $\sin t=-\tfrac{1}{2}$. In the given range the valid clean choice is $\sin t=1$, giving $t=\tfrac{\pi}{2}$.

Step 5: Find the acceleration formula.
Acceleration is the derivative of velocity: \[ a=\frac{dv}{dt}=-12\sin2t+6\cos t. \]
Step 6: Plug in $t=\tfrac{\pi}{2}$.
At $t=\tfrac{\pi}{2}$, $\sin2t=\sin\pi=0$ and $\cos t=\cos\tfrac{\pi}{2}=0$. So \[ a=-12(0)+6(0)=0. \] The acceleration is zero at that instant.
\[ \boxed{0\ \text{units/sec}^2} \]