Question

a particle moves along the x - axis so that at time t ≥ 0 its velocity is given by v(t)=-2t + 8. determine the acceleration of the particle at t = 9.

Explanation:

Step1: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Given $v(t)=- 2t + 8$.

Step2: Differentiate the velocity function

Using the power - rule for differentiation $\frac{d}{dt}(at^n)=nat^{n - 1}$, for $v(t)=-2t+8$, $a(t)=\frac{d}{dt}(-2t + 8)$.
$a(t)=\frac{d}{dt}(-2t)+\frac{d}{dt}(8)=-2+0=-2$.

Step3: Note the nature of the acceleration function

The acceleration function $a(t)=-2$ is a constant function. So, for any value of $t\geq0$, including $t = 9$, the acceleration is the same.

Answer:

$-2$