Question

the position function of an object moving along a straight line is s(t) = 1/15t^3 - 1/2t^2 + 5t^(-1). what is the objects acceleration at t = 5? 1/5 -1/5 23/25 27/25

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

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 $v(t)$)

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

Step3: Find the second - derivative (acceleration function $a(t)$)

Differentiate $v(t)$ again using the power rule. $a(t)=v^\prime(t)=\frac{1}{5}\times2t-1-5\times(-2)t^{-3}=\frac{2}{5}t - 1+\frac{10}{t^{3}}$.

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

Substitute $t = 5$ into $a(t)$: $a(5)=\frac{2}{5}\times5-1+\frac{10}{5^{3}}=2 - 1+\frac{10}{125}=1+\frac{2}{25}=\frac{25 + 2}{25}=\frac{27}{25}$.

Answer:

$\frac{27}{25}$