Question

a particle moves along the x - axis so that at time t ≥ 0 its position is given by x(t)=t² - 11t + 30. determine the acceleration of the particle at t = 4.

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position function $x(t)$. Given $x(t)=t^{2}-11t + 30$, using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=x^\prime(t)=2t-11$.

Step2: Find the acceleration function

The acceleration $a(t)$ is the derivative of the velocity function $v(t)$. Since $v(t)=2t - 11$, then $a(t)=v^\prime(t)=2$.

Step3: Evaluate the acceleration at $t = 4$

Since $a(t)$ is a constant function ($a(t)=2$ for all $t\geq0$), when $t = 4$, $a(4)=2$.

Answer:

$2$