Question

for time $t > 0$, the position of a particle moving along the $x$-axis is given by $x(t)=t^{4}-4t^{3}$. what is the acceleration of the particle when the velocity of the particle is 0? (a) -16 (b) 0 (c) 36 (d) 96

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position function $x(t)$. Given $x(t)=t^{4}-4t^{3}$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=x^\prime(t)=4t^{3}-12t^{2}$.

Step2: Find the values of $t$ when $v(t) = 0$

Set $v(t)=4t^{3}-12t^{2}=0$. Factor out $4t^{2}$: $4t^{2}(t - 3)=0$. So $t = 0$ or $t=3$. Since $t>0$, we consider $t = 3$.

Step3: Find the acceleration function

The acceleration $a(t)$ is the derivative of the velocity function. Differentiate $v(t)=4t^{3}-12t^{2}$ with respect to $t$. Using the power - rule, $a(t)=v^\prime(t)=12t^{2}-24t$.

Step4: Evaluate the acceleration at $t = 3$

Substitute $t = 3$ into the acceleration function: $a(3)=12\times3^{2}-24\times3=12\times9-72=108 - 72=36$.

Answer:

C. 36