Chapter 4, Problem 38RE

(a)

To determine

To find : The velocity.

Answer to Problem 38RE

The answer: $v(t)=-3t^2-6t+4$

Explanation of Solution

Given information:

The function is: $s(t)=3+4t-3t^2-t^3$

Calculation:

By calculating the position function, the velocity function is discovered.

  $\begin{array}{c}v(t)=s^{\prime }(t)\\ =4-6t-3t^2\end{array}$

Therefore, the required velocity is $v(t)=-3t^2-6t+4$ .

(b)

To determine

To find: The acceleration.

Answer to Problem 38RE

The answer: $a(t)=-6t-6$

Explanation of Solution

Given information:

The function is: $s(t)=3+4t-3t^2-t^3$

Calculation:

The velocity function, or the second derivative of the position function, is used to derive the acceleration function.

  $\begin{array}{c}a(t)=v\text{'}(t)\\ =-6t-6\end{array}$

Therefore, the required acceleration is $a(t)=-6t-6$ .

(c)

To determine

To describe: The motion of the particle $t\ge 0$

Answer to Problem 38RE

The answer: The particle starts at 3 and is moving in the positive direction but decelerating until a position of $4.128$ at $t\approx 0.528$

The particle then moves in the negative direction and is accelerating for $t>0.528$

Explanation of Solution

Given information:

The function is: $s(t)=3+4t-3t^2-t^3$

Calculation:

From part $(a)$ :

  $v(t)=-3t^2-6t+4$

Set, $v(t)=0$ and obtain:

  $\begin{array}{c}0=4-6t-3t^2\\ \Rightarrow 3t^2+6t=4\\ \Rightarrow t^2+2t=\frac{4}{3}\\ \Rightarrow t^2+2t+1=\frac{4}{3}+1\\ $ \Rightarrow {(t+1)}^2=\frac{7}{3}$\Rightarrow t+1=\sqrt{\frac{7}{3}}\\ \Rightarrow t=-1+\sqrt{\frac{7}{3}}\\ \Rightarrow t\approx 0.528\end{array}$

The particle is moving to the right from $t=0$ to

  $t\approx 0.528$ and then the particles moves to the left.

From part $(b)$ :

  $a(t)=-6t-6$

Set $a(t)=0$ and obtain:

  $\begin{array}{c}0=-6-6t\\ \Rightarrow 6t=-6\\ \Rightarrow t=-1\end{array}$

So,

  $\begin{array}{c}a(1)=-6-6(1)\\ =-6-6\\ =-12<0\\ $ s(0)=3+4(0)-3{(0)}^2-0^3$=3\\ $ s(0.528)=3+4(0.528)-3{(0.528)}^2-{0.528}^3$\approx 4.128\end{array}$

Therefore, the required the particle starts at 3 and is moving in the positive direction but decelerating until a position of $4.128$ at $t\approx 0.528$ .

The particle then moves in the negative direction and is accelerating for $t>0.528$ .