Question

nyamekye is driving his car in a straight line such that for a short time its velocity is defined by $v = (3t^3 + 2t)$ m/s, where t is in seconds. determine the acceleration of his car when $t = 2$ s.

Explanation:

Step1: Recall acceleration definition

Acceleration $a$ is the derivative of velocity $v$ with respect to time $t$, so $a = \frac{dv}{dt}$.

Step2: Differentiate the velocity function

Given $v = 3t^3 + 2t$, apply power rule:
$\frac{dv}{dt} = 3\times3t^{2} + 2 = 9t^2 + 2$

Step3: Substitute $t=2$ into $a(t)$

Substitute $t=2$ into $a(t)=9t^2 + 2$:
$a(2) = 9\times(2)^2 + 2 = 9\times4 + 2$

Step4: Calculate final value

Compute the numerical result:
$9\times4 + 2 = 36 + 2 = 38$

Answer:

$\boldsymbol{38\ \text{m/s}^2}$