Chapter 5, Problem 38RE

a.

To determine

To find the velocity of particle.

Answer to Problem 38RE

The velocity of particle is $v\left(t\right)=4-6t-3t^2$

Explanation of Solution

Given:

A particle is moving along a line with position function $s\left(t\right)=3+4t-3t^2-t^3$ .

Calculation:

Since , $v\left(t\right)=\frac{d}{dt}\left(s\left(t\right)\right)=\frac{d}{dt}\left(3+4t-3t^2-t^3\right)=4-6t-3t^2$

The velocity of particle is $v\left(t\right)=4-6t-3t^2$ .

b.

To determine

To find the acceleration of particle..

Answer to Problem 38RE

The acceleration of particle is $a\left(t\right)=-6-6t$

Explanation of Solution

Given:

A particle is moving along a line with position function $s\left(t\right)=3+4t-3t^2-t^3$

Calculation:

Since, acceleration of particle

  $a\left(t\right)=\frac{d^2}{d{t}^2}\left(s\left(t\right)\right)=\frac{d}{dt}\left(\frac{d}{dt}s\left(t\right)\right)=\frac{d}{dt}\left(4-6t-3t^2\right)=-6-6t$

Therefore, the acceleration of particle is $a\left(t\right)=-6-6t$ .

c.

To determine

To describe the motion of the particle for $t\ge 0$ .

Answer to Problem 38RE

when $0\le t\le 0.572$ then velocity of particle as well as its acceleration is positive it means that particle is accelerating (speeding up) and when $t\ge 0.572$ velocity of particle is negative and its acceleration is positive means particle is deaccelerating (its speed is decreasing).

Explanation of Solution

Given:

A particle is moving along a line with position function $s\left(t\right)=3+4t-3t^2-t^3$ .

Calculation:

Initial position of particle is at $t=0$

  $\begin{array}{l}s\left(t\right)=3+4t-3t^2-t^3\\ s\left(0\right)=3+4\left(0\right)-3{\left(0\right)}^2-{\left(0\right)}^3\\ s\left(0\right)=3\end{array}$

Below is the graph of $v\left(t\right)=4-6t-3t^2$

  

From above graph it can be concluded that when $0\le t\le 0.572$ then $v\left(t\right)\ge 0$

And when $t\ge 0.572$ then $v\left(t\right)\le 0$ .

Now, at $t=0.572$

  $\begin{array}{l}s\left(t\right)=3+4t-3t^2-t^3\\ s\left(0.572\right)=3+4\left(0.572\right)-3{\left(0.572\right)}^2-{\left(0.572\right)}^3=4.11\end{array}$

Below is the graph of $a\left(t\right)=-6-6t$

  

From above graph it can be concluded that for $t\ge 0$ acceleration of particle is positive.

therefore, when $0\le t\le 0.572$ then velocity of particle as well as its acceleration is positive it means that particle is accelerating (speeding up) and when $t\ge 0.572$ velocity of particle is negative and its acceleration is positive means particle is deaccelerating (its speed is decreasing).