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)…

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) c) the velocity and acceleration when t = 4 sec a) v(t)= b) a(t)= c) when t = 4 sec, the velocity is (simplify your answer.) when t = 4 sec, the acceleration is (simplify your answer.)

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) c) the velocity and acceleration when t = 4 sec a) v(t)= b) a(t)= c) when t = 4 sec, the velocity is (simplify your answer.) when t = 4 sec, the acceleration is (simplify your answer.)

Answer

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v(t)$ is the derivative of displacement $s(t)$. Given $s(t)=4t^{2}+4t$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(4t^{2}+4t)$. $v(t)=4\times2t+4=8t + 4$.

Step2: Recall acceleration - velocity relation

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

Step3: Find velocity at $t = 4$

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

Step4: Find acceleration at $t = 4$

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

Answer:

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