Question

question
the position of a car traveling along a highway is given by the function s(t)=2t^4 - 9t^3 - 6t^2 - 8 where t is measured in seconds and s is measured in meters. find the acceleration of the car at t = 2 seconds.
provide your answer below:
□ m/s²

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

Velocity $v(t)$ is the first - derivative of position $s(t)$, and acceleration $a(t)$ is the second - derivative of position $s(t)$. First, find the first - derivative of $s(t)=2t^{4}-9t^{3}-6t^{2}-8$.
$v(t)=s^\prime(t)=\frac{d}{dt}(2t^{4}-9t^{3}-6t^{2}-8)$
Using the power rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=8t^{3}-27t^{2}-12t$.

Step2: Find the second - derivative of $s(t)$

Acceleration $a(t)=v^\prime(t)=s^{\prime\prime}(t)$. Differentiate $v(t)=8t^{3}-27t^{2}-12t$ with respect to $t$.
$a(t)=\frac{d}{dt}(8t^{3}-27t^{2}-12t)$
Using the power rule again, $a(t)=24t^{2}-54t - 12$.

Step3: Evaluate the acceleration at $t = 2$

Substitute $t = 2$ into the acceleration function $a(t)$.
$a(2)=24\times2^{2}-54\times2 - 12$
$a(2)=24\times4-108 - 12$
$a(2)=96-108 - 12$
$a(2)=96-(108 + 12)$
$a(2)=96 - 120=-24$.

Answer:

$-24$