Chapter 2.4, Problem 37E

a.

To determine

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

Answer to Problem 37E

In the beginning, the particle was moving in a positive direction until $t\approx 1.153$ , then it switched to moving in a negative direction until $t\approx 3.180$ , and then it switched back to moving in a positive direction again.

Explanation of Solution

Given:

The position $x-$ co-ordinate of a particle moving on the line: $y=2$

And $x\left(t\right)=2t^3-13t^2+22t-5$ .

Using the power rule, the derivative of $x\left(t\right)$ can be identified.

This derivative will represent velocity:

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(2t^3-13t^2+22t-5\right)\\ =6t^2-26t+22\end{array}$

Find the velocity when it equal to $0$:

As the particle will have zero velocity at maximum or minimum, this will tell us the time when the particle reached its maximum/minimum.

Use the quadratic formula:

  $\begin{array}{c}6t^2-26t+22=0\\ t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ t_{1,2}=\frac{26\pm \sqrt{{\left(-26\right)}^2-4\times 6\times 22}}{2\times 6}\\ t_{1,2}=\frac{26\pm \sqrt{148}}{12}\end{array}$

Therefore, until approximately $t\approx 1.153$ , the particle was moving in a positive direction.

Once the particle's velocity was $0$ , it changed direction and started moving in a negative direction until $t\approx 3.180$ , when it was again $0$ .

From that point on, the particle again changed direction and moved in a positive direction.

b.

To determine

To Describe: The time, in which the particle speeds up and slows down.

Answer to Problem 37E

At the intervals $1.153\le t\le 2.167$ and $t\ge 3.18$ the particle is speeding up.

At the intervals $0\le t\le 1.153$ and $2.167\le t\le 3.18$ the particle is slowing down.

Explanation of Solution

Given:

The position $x-$ co-ordinate of a particle moving on the line: $y=2$

Given the equation: $x\left(t\right)=2t^3-13t^2+22t-5$ .

When velocity and acceleration have the same sign (both positive or both negative), a particle is speeding up, while a particle is slowing down when the signs are opposite (one is positive and the other is negative).

From part a) The velocity is positive when $0\le t\le 1.153$ and $t\ge 3.18$ and velocity is negative when $1.153\le t\le 3.18$ .

Find the second derivative of $x\left(t\right)$ to find when the acceleration is positive and negative,

  $\begin{array}{c}x\left(t\right)=2t^3-13t^2+22t-5\\ x^{\prime }\left(t\right)=6t^2-26t+22\\ x^{\prime \prime }\left(t\right)=12t-26\end{array}$

Set the above equation equal to $0$ :

  $\begin{array}{c}12t-26=0\\ 12t=26\\ t\approx 2.167\end{array}$

Test the value between 0 and $2.167$ :

So, Take $t=1$ :

  $\begin{array}{c}x\text{'}\text{'}\left(1\right)=12\left(1\right)-26\\ =12-26\\ =-14<0\end{array}$

Test the value greater than $2.167$ :;

So, take $t=3$ :

  $\begin{array}{c}x\text{'}\text{'}\left(3\right)=12\left(3\right)-26\\ =36-26\\ =10>0\end{array}$

Hence, the acceleration is positive when $t\ge 2.167$ and negative when $0\le t\le 2.167$ .

At the interval, $0\le t\le 1.153$ , acceleration is negative, and velocity is positive.

At the interval, $1.153\le t\le 2.167$ , acceleration and velocity are both negative.

At the interval, $2.167\le t\le 3.18$ acceleration is positive, and velocity is negative.

Both acceleration and velocity are positive for $t\ge 3.18$ .

So,

At the intervals $1.153\le t\le 2.167$ and $t\ge 3.18$ the particle is speeding up.

At the intervals $0\le t\le 1.153$ and $2.167\le t\le 3.18$ the particle is slowing down.

c.

To determine

To Find: The time the particle changing its direction.

Answer to Problem 37E

The particle changing its direction at $t=1.153$ and $t=3.180$ .

Explanation of Solution

Given:

The position $x-$ co-ordinate of a particle moving on the line: $y=2$

The equation: $x\left(t\right)=2t^3-13t^2+22t-5$ .

When the particle velocity is $0$ , the particle changes its direction.

As mentioned in part $\left(a\right)$ , for $t=1.153$ and $t=3.180$ , the velocity is zero.

d.

To determine

To Find: The time at which the particle is at rest.

Answer to Problem 37E

At $t\approx 1.153$and $t\approx 3.180$ , the particle is at rest.

Explanation of Solution

Given:

The position $x-$ co-ordinate of a particle moving on the line: $y=2$

The equation: $x\left(t\right)=2t^3-13t^2+22t-5$ .

When the particle velocity is $0$ , the particle is at rest.

From part $\left(a\right)$ its calculated that the velocity will be equal to $0$ at $t\approx 1.153$and $t\approx 3.180$ .

So, at $t\approx 1.153$and $t\approx 3.180$ , the particle is at rest.

e.

To determine

To Describe: The velocity and the speed of the particle.

Answer to Problem 37E

Velocity:

Initially, the velocity was positive, but it stopped increasing, and then it became negative, and then it increased again, and then it became positive again then continue to increase.

Speed:

During the first part of the equation, the speed drops to zero at $t\approx 1.15$ , then increases until $t\approx 2.17$ , decreases again until $t=3.18$ when it $0$ again, then increases again.

Explanation of Solution

Given:

The position $x-$ co-ordinate of a particle moving on the line: $y=2$ .

The equation: $x\left(t\right)=2t^3-13t^2+22t-5$ .

State the velocity and the speed of the particle:

It's calculates from part $\left(d\right)$ the particle is at rest at $t=1.153$ and $t=3.180$ .

Hence at the time at $t=1.153$ and $t=3.180$ the velocity and the speed will be $0$ .

There is a positive velocity at first since,

  $\begin{array}{c}v\left(0\right)=6\times 0^2-26\cdot 0+22\\ =22\end{array}$

At $t=0$ , it will gradually decrease until it becomes negative.

Thereafter, it will increase to infinity again.

Due to the initial decrease in velocity, the speed will also decrease until $t=1.153$ .

It then increases until $t=2.17$ (from part $\left(b\right)$ ) and then decreases until $t=3.180$ . Thereafter, it will increase steadily.

Diagrammatic representation is as follows:

The graph of $x\left(t\right)$ represented by green curve and $x^{\prime }\left(t\right)$ represented by blue curve.

  

e.

To determine

To Describe: The time in which the particle at the point $\left(5,2\right)$ .

Answer to Problem 37E

The time in which the particle at the point $\left(5,2\right)$ are $t=0.745,1.626,4.129$ .

Explanation of Solution

Given:

The position $x-$ co-ordinate of a particle moving on the line: $y=2$ .

Given the equation: $x\left(t\right)=2t^3-13t^2+22t-5$ .

Find the time, the particle is at $\left(5,2\right)$ :

Position of the particle at time $t$ is $\left(x\left(t\right),2\right)$ . The given equation for $x\left(t\right)$ should be equal to $5$ since the position is at $\left(5,2\right)$ :

So, set $x\left(t\right)=5$ . Given that $x\left(t\right)=2t^3-13t^2+22t-5$ .

  $2t^3-13t^2+22t-5=5$

Subtract $5$ from both sides:

  $\begin{array}{c}2t^3-13t^2+22t-5-5=5-5\\ 2t^3-13t^2+22t-10=0\end{array}$

Find the value of $t$ by using the graph:

  

Thus, the values of $t$ are $t=0.745,1.626,4.129$ .