Question

the position function of an object moving along a straight line is s(t) = 1/3t^3 - 1/2t^2 + 5t - 1. what is the objects acceleration at t = 5? question 7 of 10

Explanation:

Step1: Recall the relationship between position and acceleration

Acceleration $a(t)$ is the second - derivative of position function $s(t)$. Given $s(t)=\frac{1}{15}t^{3}-\frac{1}{2}t^{2}+5t - 1$.

Step2: Find the first - derivative of $s(t)$

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 of $s(t)$

Differentiate $v(t)$ with respect to $t$. $a(t)=v^\prime(t)=s^{\prime\prime}(t)=\frac{1}{5}\times2t-1=\frac{2}{5}t - 1$.

Step4: Evaluate $a(t)$ at $t = 5$

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

Answer:

$1$