Question

a particle moves along the x - axis. its position is given by p(t)=2t^{3}-8t^{2}+15 inches where time is measured in seconds. what is the objects acceleration at t = 3 seconds? attempt 1: 10 attempts remaining.

Explanation:

Step1: Recall the relationship between position and acceleration

The acceleration $a(t)$ is the second - derivative of the position function $p(t)$. Given $p(t)=2t^{3}-8t^{2}+15$.

Step2: Find the first - derivative (velocity function)

Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=p^\prime(t)=\frac{d}{dt}(2t^{3}-8t^{2}+15)=6t^{2}-16t$.

Step3: Find the second - derivative (acceleration function)

Differentiate $v(t)$ with respect to $t$. $a(t)=v^\prime(t)=p^{\prime\prime}(t)=\frac{d}{dt}(6t^{2}-16t)=12t - 16$.

Step4: Evaluate the acceleration at $t = 3$

Substitute $t = 3$ into $a(t)$. $a(3)=12\times3-16$.
$a(3)=36 - 16=20$ inches per second squared.

Answer:

20 inches per second squared