question 15 5 pts solve the problem. if s is a distance given by s(t)=2t^4 + 4t^3 + 4t, find the…

question 15 5 pts solve the problem. if s is a distance given by s(t)=2t^4 + 4t^3 + 4t, find the acceleration, a(t). edit view insert format tools table 12pt paragraph b i u a t^2

question 15 5 pts solve the problem. if s is a distance given by s(t)=2t^4 + 4t^3 + 4t, find the acceleration, a(t). edit view insert format tools table 12pt paragraph b i u a t^2

Answer

Explanation:

Step1: Recall the relationship between distance, velocity and acceleration

Acceleration is the second - derivative of distance with respect to time. First, find the velocity $v(t)$ which is the first - derivative of $s(t)$. $v(t)=s^\prime(t)=\frac{d}{dt}(2t^{4}+4t^{3}+4t)$

Step2: Differentiate $s(t)$ to get $v(t)$

Using the power rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=2\times4t^{3}+4\times3t^{2}+4=8t^{3}+12t^{2}+4$.

Step3: Differentiate $v(t)$ to get $a(t)$

$a(t)=v^\prime(t)=\frac{d}{dt}(8t^{3}+12t^{2}+4)$

Step4: Calculate the second - derivative

Again using the power rule, $a(t)=8\times3t^{2}+12\times2t+0 = 24t^{2}+24t$.

Answer:

$a(t)=24t^{2}+24t$