Question

a particle moves along the x - axis so that at time $tgeq0$ its position is given by $x(t)=t^{3}-11t^{2}+39t$. determine the velocity of the particle at $t = 4$.

Explanation:

Step1: Recall velocity - position relationship

The velocity function $v(t)$ is the derivative of the position function $x(t)$. So, if $x(t)=t^{3}-11t^{2}+39t$, then $v(t)=x^\prime(t)$.
Using the power - rule for differentiation $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(t^{3}-11t^{2}+39t)=3t^{2}-22t + 39$.

Step2: Evaluate velocity at $t = 4$

Substitute $t = 4$ into the velocity function $v(t)$.
$v(4)=3(4)^{2}-22(4)+39$.
First, calculate $3(4)^{2}=3\times16 = 48$, $22(4)=88$.
Then $v(4)=48-88 + 39$.
$v(4)=48+39-88= - 1$.

Answer:

$-1$