6 mark for review a particle moves along the x - axis so that at time t ≥ 0 its position is given by…

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position function $x(t)$. Given $x(t)=2t^{3}-9t^{2}-60t + 4$, then $v(t)=x^\prime(t)=6t^{2}-18t - 60=6(t^{2}-3t - 10)=6(t - 5)(t+2)$.

Step2: Find the critical - points

Set $v(t)=0$. Since $v(t)=6(t - 5)(t + 2)=0$, the critical - points are $t=-2$ and $t = 5$. We only consider $t = 5$ in the interval $0\leq t\leq7$.

Step3: Evaluate the position function at the critical - point and endpoints

$x(0)=2(0)^{3}-9(0)^{2}-60(0)+4 = 4$. $x(5)=2(5)^{3}-9(5)^{2}-60(5)+4=2\times125-9\times25-300 + 4=250-225-300 + 4=-271$. $x(7)=2(7)^{3}-9(7)^{2}-60(7)+4=2\times343-9\times49-420 + 4=686-441-420 + 4=-171$.

Step4: Calculate the total distance

The total distance $d$ is given by $d=\vert x(5)-x(0)\vert+\vert x(7)-x(5)\vert$. $\vert x(5)-x(0)\vert=\vert-271 - 4\vert=\vert-275\vert = 275$. $\vert x(7)-x(5)\vert=\vert-171+271\vert = 100$. $d = 275+100=375$.

Answer:

D. 375