Chapter 6.4, Problem 58E

(a)

To determine

Particle’s velocity at time $t=5$ .

Answer to Problem 58E

The velocity at $t=5$ is $2\text{m}/\text{s}$ .

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

The velocity of s is s’.

Since

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Then

According to FTC (Fundamental Theorem of Calculus),

  $s\text{'}\left(t\right)=\frac{d}{dt}\left({\displaystyle {\int }_0^{t}f\left(x\right)}dx\right)=f\left(t\right)$

The velocity at $t=5$ is $s\text{'}\left(5\right)$ .

From the graph,

  $s\text{'}\left(5\right)=f\left(5\right)=2$

Thus,

The velocity at $t=5$ is $2\text{m}/\text{s}$ .

(b)

To determine

Whether the acceleration of particle at time $t=5$ positive or negative.

Answer to Problem 58E

The acceleration of particle is negative at $t=5$ .

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

The acceleration of s is s”.

Since

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Then

According to FTC (Fundamental Theorem of Calculus),

  $s\text{'}\left(t\right)=\frac{d}{dt}\left({\displaystyle {\int }_0^{t}f\left(x\right)}dx\right)=f\left(t\right)$

Second derivative:

  $s\text{'}\text{'}\left(t\right)=\frac{d}{dt}\left[f\left(t\right)\right]=f\text{'}\left(t\right)$

That means

  $s\text{'}\text{'}\left(5\right)=f\text{'}\left(5\right)$

From the graph,

The slope of f at 5 is negative.

Thus,

  $s\text{'}\text{'}\left(5\right)$ is negative.

Therefore,

The acceleration of particle is negative at $t=5$ .

(c)

To determine

Particle’s position at time $t=3$ .

Answer to Problem 58E

Particle’s position at 3 seconds is 4.5.

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

The region between $x=0$ and $x=3$ is a right angle triangle, with base of 3 and the height of 3.

Then

The position at 3 seconds:

  $s\left(3\right)={\displaystyle {\int }_0^{3}f\left(x\right)dx=\frac{1}{2}}\times 3\times 3=\frac{9}{2}=4.5$

Therefore,

Particle’s position at 3 seconds is 4.5.

(d)

To determine

Time during the first 9 sec for s have its largest value.

Answer to Problem 58E

The largest value for s occurs at $t=6$ .

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

s will have the largest value when it reaches the absolute maximum.

In order to know where the maximum occurs,

We are required to find at what value of t the first derivative of s switches from being positive to negative.

Since

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Then

According to FTC (Fundamental Theorem of Calculus),

  $s\text{'}\left(t\right)=\frac{d}{dt}\left({\displaystyle {\int }_0^{t}f\left(x\right)}dx\right)=f\left(t\right)$

From the graph,

f is positive for 0 to 6 and negative for 6 to 9.

Therefore,

  $t=6$ is the location of a maximum.

For $t=6$ to be absolute maximum and not a local maximum,

s must be larger at $t=6$ then it is at its endpoints.

Since

  $s\text{'}>0$ for $0<t<6$ ,

Then

  $s\left(0\right)<s\left(6\right)$

Since

  $s\text{'}<0$ for $6<t<9$ ,

Then

  $s\left(6\right)>s\left(9\right)$

Thus,

s is larger at $t=6$ at its endpoints.

Therefore,

The largest value for s occurs at $t=6$ .

(e)

To determine

Approximate the zero value of acceleration.

Answer to Problem 58E

The acceleration is zero when $t=4$ and $t=7$ .

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

Since

At $t=4$

We have

  $s\text{'}\text{'}\left(t\right)=f\text{'}\left(t\right)=0$

Also,

At $t=7$

We have

  $s\text{'}\text{'}\left(t\right)=f\text{'}\left(t\right)=0$

Therefore,

Acceleration is zero at both 4 seconds and 7 seconds.

(f)

To determine

Movement of the particle towards and away from the origin.

Answer to Problem 58E

The particle moves away from the origin in the positive direction on the interval [0, 6].

The particle moves towards the origin in the negative direction on the interval [6, 9].

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

Note that

  $s\left(0\right)=0$

Such that

The particle is at origin at $t=0$ .

On the interval [0, 6]:

We have

  $s\text{'}\left(t\right)=f\left(t\right)>0$

Thus,

The particle moves away from the origin in the positive direction.

On the interval [6, 9]:

We have

  $s\text{'}\left(t\right)=f\left(t\right)<0$

Also,

The area below the x-axis is smaller above the x-axis.

Thus,

The particle moves towards the origin in negative direction.

(g)

To determine

Side of the origin for the particle at time $t=9$ .

Answer to Problem 58E

The particle lies on positive side of the origin at time $t=9$ .

Explanation of Solution

Given information:

Position at time t (sec) of a particle moving along a coordinate axis:

  $s={\displaystyle {\int }_0^{t}f\left(x\right)}dx$

Where,

f is the differentiable function

  

Since the particle starts at the origin,

Then

  $s\left(0\right)={\displaystyle {\int }_0^{0}f\left(x\right)}dx=0$

Also,

The area below the x-axis between $x=6$ and $x=9$ is smaller than the area above x-axis between $x=0$ and $x=6$ .

Thus,

  $s\left(9\right)={\displaystyle {\int }_0^{9}f\left(x\right)}dx>0$

Therefore,

The particle lies on the positive side of the origin at 9 seconds.