given s(t)=4t² + 4t, where s(t) is in feet and t is in seconds, find each of the following. a) v(t) b) a(t)…

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v(t)$ is the derivative of displacement $s(t)$. Given $s(t)=4t^{2}+4t$, use the power - rule for differentiation $\frac{d}{dt}(t^{n}) = nt^{n - 1}$. $v(t)=\frac{d}{dt}(4t^{2}+4t)$

Step2: Apply the sum - rule and power - rule of differentiation

The sum - rule states that $\frac{d}{dt}(u + v)=\frac{d}{dt}(u)+\frac{d}{dt}(v)$. Here $u = 4t^{2}$ and $v = 4t$. $v(t)=\frac{d}{dt}(4t^{2})+\frac{d}{dt}(4t)$ $v(t)=4\times2t+4\times1$ $v(t)=8t + 4$

Step3: Recall acceleration - velocity relation

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Since $v(t)=8t + 4$, using the power - rule for differentiation. $a(t)=\frac{d}{dt}(8t + 4)$

Step4: Apply the sum - rule and power - rule of differentiation

$a(t)=\frac{d}{dt}(8t)+\frac{d}{dt}(4)$ $a(t)=8+0=8$

Step5: Find velocity at $t = 4$

Substitute $t = 4$ into $v(t)$. $v(4)=8\times4+4$ $v(4)=32 + 4=36$

Step6: Find acceleration at $t = 4$

Since $a(t)=8$ (constant), $a(4)=8$

Answer:

a) $v(t)=8t + 4$ b) $a(t)=8$ c) $v(4)=36$ feet per second, $a(4)=8$ feet per second squared