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: Find velocity function

Velocity $v(t)$ is derivative of position - function $s(t)$. Using power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, for $s(t)=4t^{2}+4t$, we have $v(t)=\frac{d}{dt}(4t^{2}+4t)=4\times2t^{2 - 1}+4\times1t^{1 - 1}=8t + 4$.

Step2: Find acceleration function

Acceleration $a(t)$ is derivative of velocity function $v(t)$. Since $v(t)=8t + 4$, then $a(t)=\frac{d}{dt}(8t + 4)=8$.

Step3: Find velocity at $t = 4$

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

Step4: Find acceleration at $t = 4$

Since $a(t)=8$ (constant), when $t = 4$, $a(4)=8$ ft/sec².

Answer:

a) $v(t)=8t + 4$ b) $a(t)=8$ c) When $t = 4$ sec, the velocity is 36 ft/sec. When $t = 4$ sec, the acceleration is 8 ft/sec².