Chapter 3.4, Problem 37E

a.

To determine

To describe:Themotion of the particle.

Explanation of Solution

Given information:

1. Particle is moving on line $y=2$ .
2. Function defining change in $x$ coordinate in terms of time $t$ is $x\left(t\right)=2t^3-13t^2+22t-5$

As particle is moving on $y=2$ , y coordinate of the particle will always be equal to 2.

Consider the graph of function $x=f\left(t\right)=2t^3-13t^2+22t-5$ governing value of $x$ coordinate with respect to time $t$ .

  

On analyzing the graph, at $t=0$ that is starting point value of $x$ coordinate is $-5.$

That is position of the particle on line $y=2$ at $t=0$ is $\left(-5,2\right)$ .

Then value of $x$ coordinate increases to $6.619$ at $t=1.153$ seconds.

That is position of the particle on line $y=2$ at $t=1.153$ seconds is $\left(6.169,2\right)$ .

Then value of $x$ coordinate decreases to $-2.186$ at $t=3.18$ seconds.

That is position of the particle on line $y=2$ at $t=3.18$ seconds is $\left(-2.186,2\right)$ .

And after $t=3.18$ seconds value of $x$ coordinate increases with increase in time .

Describe the motion:

Base on above fact starting point of the motion of particle is $\left(-5,2\right)$ , then particle moves on positive direction of $x$ till point $\left(6.169,2\right)$ at $t=1.153$ seconds . Then it changes direction and move towards the negative direction of $x$ axis till point $\left(-2.186,2\right)$ at $t=3.18$ . After $t=3.18$ , particle again changes its direction and start moving in positive direction of $x$ and keep moving with increasing speed along line $y=2$ . Increase in speed can be concluded base of the fact that cure of $f\left(t\right)$ becomes more steeper as time increases .

b.

To determine

To find:The time interval when particle speeds up and slows down.

Answer to Problem 37E

Particle speeds up for time interval $t=\left(0,0.268\right)\cup \left(1.153,2.167\right)\cup \left(3.18,\infty \right)$ .

Explanation of Solution

Given information:

1. Particle is moving on line $y=2$ .
2. Function defining change in $x$ coordinate in terms of time $t$ is $x\left(t\right)=2t^3-13t^2+22t-5$

Calculation:

As particle is moving on $y=2$ , y coordinate of the particle will always be equal to 2.

Consider the graph of function $x=f\left(t\right)=2t^3-13t^2+22t-5$ governing value of $x$ coordinate with respect to time $t$ .

  

On analyzing the graph , on line $y=2$ , at starting that is $A$ , $t=0$ and particle was at rest . then speeds up till $B$ where $t=0.268$ seconds . then slows down and come to rest at $C$ to change its direction. Then again speeds up till $D$ at $t=2.167$ .After that slowdown to come at rest to chage the direction at $E$ when $t=3.18$ seconds and after than speeds up till end.

Based on above analysis , particle speeds up for time interval $\underset{\_}{t=\left(0,0.268\right)\cup \left(1.153,2.167\right)\cup \left(3.18,\infty \right)}$ and slows down for interval $\underset{\_}{t=\left(0.268,1.153\right)\cup \left(2.167,3.18\right)}$ .

c.

To determine

To find:The time interval when particle changesthe direction.

Answer to Problem 37E

Particle changes direction at $t\approx 1.153\sec$ and $\underset{\_}{t\approx 3.18\sec }$ .

Explanation of Solution

Given information:

1. Particle is moving on line $y=2$ .
2. Function defining change in $x$ coordinate in terms of time $t$ is $x\left(t\right)=2t^3-13t^2+22t-5$

Calculation:

As particle is moving on $y=2$ , y coordinate of the particle will always be equal to 2.

Consider the graph of function $x=f\left(t\right)=2t^3-13t^2+22t-5$ governing value of $x$ coordinate with respect to time $t$ .

  

On analyzing the graph , on line $y=2$ , at starting that is $A$ , $t=0$ and particle was at rest . then speeds up till $B$ where $t=0.268$ seconds . then slows down and come to rest at $C$ to change its direction at $t=1.153$ . Then again speeds up till $D$ at $t=2.167$ .After that slowdown to come at rest to change the direction at $E$ when $t=3.18$ seconds and after than speeds up till end.

Based on above analysis , particle changes direction at $\underset{\_}{t\approx 1.153\sec }$ and $\underset{\_}{t\approx 3.18\sec }$ .

d.

To determine

To find:The time interval when particlewas at rest.

Answer to Problem 37E

Particle was at rest at $t\approx 1.153\sec$ and $t\approx 3.18\sec$ .

Explanation of Solution

Given information:

1. Particle is moving on line $y=2$ .
2. Function defining change in $x$ coordinate in terms of time $t$ is $x\left(t\right)=2t^3-13t^2+22t-5$

Calculation:

As particle is moving on $y=2$ , y coordinate of the particle will always be equal to 2.

Consider the graph of function $x=f\left(t\right)=2t^3-13t^2+22t-5$ governing value of $x$ coordinate with respect to time $t$ .

  

On analyzing the graph , on line $y=2$ , at starting that is $A$ , $t=0$ and particle was at rest . then speeds up till $B$ where $t=0.268$ seconds . then slows down and come to rest at $C$ to change its direction at $t=1.153$ . Then again speeds up till $D$ at $t=2.167$ .After that slowdown to come at rest to change the direction at $E$ when $t=3.18$ seconds and after than speeds up till end.

Based on above analysis , particle was at rest at $\underset{\_}{t\approx 1.153\sec }$ and $\underset{\_}{t\approx 3.18\sec }$ .

e.

To determine

To describe:The velocity and speed of particle.

Explanation of Solution

Given information:

1. Particle is moving on line $y=2$ .
2. Function defining change in $x$ coordinate in terms of time $t$ is $x\left(t\right)=2t^3-13t^2+22t-5$

As particle is moving on $y=2$ , y coordinate of the particle will always be equal to 2.

Consider the graph of function $x=f\left(t\right)=2t^3-13t^2+22t-5$ governing value of $x$ coordinate with respect to time $t$ .

  

On analyzing the graph , on line $y=2$ , at starting that is $A$ , $t=0$ and particle was at rest . then speeds up till $B$ where $t=0.268$ seconds . then slows down and come to rest at $C$ to change its direction. Then again speeds up till $D$ at $t=2.167$ .After that slowdown to come at rest to chage the direction at $E$ when $t=3.18$ seconds and after than speeds up till end.

Based on above analysis , particle speeds increases for time interval $\underset{\_}{t=\left(0,0.268\right)\cup \left(1.153,2.167\right)\cup \left(3.18,\infty \right)}$ and decreases for interval $\underset{\_}{t=\left(0.268,1.153\right)\cup \left(2.167,3.18\right)}$ .

As velocity also consider direction , velocity is positive from $\underset{\_}{t=\left(0,1.153\right)\cup \left(3.18,\infty \right)}$ and was negative for $\underset{\_}{t=\left(1.153,3.18\right)}$

f.

To determine

To calculate:The time for specific position of particle.

Answer to Problem 37E

There are three times when position of the particle will be $\left(5,2\right)$ that are $\underset{\_}{t=0.745\sec }$ , $\underset{\_}{t=1.626\sec }$ and $\underset{\_}{t=4.129\sec }$ .

Explanation of Solution

Given information:

1. Particle is moving on line $y=2$ .
2. Function defining change in $x$ coordinate in terms of time $t$ is $x\left(t\right)=2t^3-13t^2+22t-5$

Calculation:

As particle is moving on $y=2$ , y coordinate of the particle will always be equal to 2.

Consider the graph of function $x=f\left(t\right)=2t^3-13t^2+22t-5$ governing value of $x$ coordinate with respect to time $t$ .

  

On analyzing the graph of motion of particle on line $y=2$ , there are three times when value of $x\left(t\right)$ is $5$ . the three times are $t=0.745$ , $t=1.626$ and $t=4.129$ and as motion of the particle is along line $y=2$ , $y$ coordinate of motion of particle will always be $2$ .

Hence it can be concluded that there are three times when position of the particle will be $\left(5,2\right)$ that are $\underset{\_}{t=0.745\sec }$ , $\underset{\_}{t=1.626\sec }$ and $\underset{\_}{t=4.129\sec }$