Question

what is the objects acceleration at t = 5? the position function of an object moving along a straight line is s(t)=\frac{1}{15}t^{3}-\frac{1}{2}t^{2}+5t - 1

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

The velocity $v(t)$ is the first - derivative of the position function $s(t)$, and the acceleration $a(t)$ is the second - derivative of the position function $s(t)$. Given $s(t)=\frac{1}{15}t^{3}-\frac{1}{2}t^{2}+5t - 1$.

Step2: Find the first - derivative (velocity function)

Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=\frac{1}{15}\times3t^{2}-\frac{1}{2}\times2t + 5=\frac{1}{5}t^{2}-t + 5$.

Step3: Find the second - derivative (acceleration function)

Differentiate $v(t)$ with respect to $t$ again. Using the power rule, $a(t)=v^\prime(t)=s^{\prime\prime}(t)=\frac{2}{5}t-1$.

Step4: Evaluate the acceleration at $t = 5$

Substitute $t = 5$ into the acceleration function $a(t)$. $a(5)=\frac{2}{5}\times5-1$.
$a(5)=2 - 1=1$.

Answer:

$1$