Question

a particle moves along a straight line and its position at time t is given by s(t)=2t^3 - 21t^2 + 36t where s is measured in feet and t in seconds. find the velocity (in ft/sec) of the particle at time t = 0: the particle stops moving (i.e. is in a rest) twice, first when t = , and again when t = what is the position of the particle at time 14? finally, what is the total distance the particle travels between time 0 and time 14? question help: video message instructor submit question

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=2t^{3}-21t^{2}+36t$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=s^\prime(t)=6t^{2}-42t + 36$.

Step2: Find the velocity at $t = 0$

Substitute $t = 0$ into $v(t)$: $v(0)=6(0)^{2}-42(0)+36=36$.

Step3: Find when the particle stops moving

The particle stops moving when $v(t)=0$. So we set $6t^{2}-42t + 36 = 0$. Divide through by 6: $t^{2}-7t + 6=0$. Factor the quadratic equation: $(t - 1)(t - 6)=0$. So $t = 1$ or $t = 6$. The first time is $t = 1$.

Step4: Find the position at $t = 14$

Substitute $t = 14$ into $s(t)$: $s(14)=2(14)^{3}-21(14)^{2}+36(14)=2\times2744-21\times196 + 36\times14=5488-4116+504=1876$.

Step5: Find the total distance traveled

We need to consider the intervals where the velocity changes sign. The critical points of $v(t)$ are $t = 1$ and $t = 6$.

• On the interval $[0,1]$, $v(t)>0$, and $s(1)=2(1)^{3}-21(1)^{2}+36(1)=2 - 21 + 36 = 17$, $s(0)=0$. The distance traveled on $[0,1]$ is $s(1)-s(0)=17$.
• On the interval $[1,6]$, $v(t)<0$, $s(6)=2(6)^{3}-21(6)^{2}+36(6)=432-756 + 216=-108$. The distance traveled on $[1,6]$ is $|s(6)-s(1)|=| - 108 - 17|=125$.
• On the interval $[6,14]$, $v(t)>0$, $s(14)=1876$. The distance traveled on $[6,14]$ is $s(14)-s(6)=1876+108 = 1984$.

The total distance is $17+125+1984=2126$.

Answer:

• Velocity at $t = 0$: 36
• First time particle stops: 1
• Second time particle stops: 6
• Position at $t = 14$: 1876
• Total distance from $t = 0$ to $t = 14$: 2126