Question

a particle moves along the x - axis so that at time t ≥ 0 its position is given by x(t)=2t³ - 9t² - 60t + 4. what is the total distance traveled by the particle over the time interval 0 ≤ t ≤ 7? a 168 b 171 c 175 d 375

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 $v(t)=x^\prime(t)=6t^{2}-18t - 60$.

Step2: Find the critical points of $v(t)$

Set $v(t)=0$, so $6t^{2}-18t - 60 = 0$. Divide through by 6: $t^{2}-3t - 10=0$. Factor: $(t - 5)(t+ 2)=0$. The critical points are $t = 5$ and $t=-2$. We only consider $t = 5$ since $t\geq0$.

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 $|x(5)-x(0)|+|x(7)-x(5)|$.
$|x(5)-x(0)|=|-271 - 4|=275$.
$|x(7)-x(5)|=|-171+271| = 100$.
$d=275+100=375$.

Answer:

D. 375