Question

a particle moves along the x - axis so that at time $tgeq0$ its position is given by $x(t)=-t^{2}+7t - 38$. determine the velocity of the particle at $t = 10$.

Explanation:

Step1: Recall velocity - position relationship

The velocity function $v(t)$ is the derivative of the position function $x(t)$. Given $x(t)=-t^{2}+7t - 38$, by the power - rule of differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we find $v(t)$.
$v(t)=\frac{d}{dt}(-t^{2}+7t - 38)=-2t + 7$

Step2: Evaluate velocity at $t = 10$

Substitute $t = 10$ into the velocity function $v(t)$.
$v(10)=-2\times10+7$
$v(10)=-20 + 7=-13$

Answer:

$-13$