Question

the position of a car traveling along a highway is given by the function s(t)=2t^4 - 10t^3 - 5t^2 + 9 where t is measured in seconds and s is measured in meters. find the acceleration of the car at t = 2 seconds. sorry, thats incorrect. try again? -24 m/s^2

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the first - derivative of position $s(t)$. So, $v(t)=s^\prime(t)$. Using the power rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have $s(t)=2t^{4}-10t^{3}-5t^{2}+9$, then $v(t)=\frac{d}{dt}(2t^{4}-10t^{3}-5t^{2}+9)=8t^{3}-30t^{2}-10t$.

Step2: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the first - derivative of velocity $v(t)$ or the second - derivative of position $s(t)$. So, $a(t)=v^\prime(t)=s^{\prime\prime}(t)$. Differentiating $v(t)=8t^{3}-30t^{2}-10t$ using the power rule, we get $a(t)=\frac{d}{dt}(8t^{3}-30t^{2}-10t)=24t^{2}-60t - 10$.

Step3: Evaluate acceleration at $t = 2$

Substitute $t = 2$ into the acceleration function $a(t)$. $a(2)=24\times2^{2}-60\times2 - 10=24\times4-120 - 10=96-120 - 10=-34$.

Answer:

$-34$ m/s²