Chapter 6.4, Problem 59E

(a)

To determine

Particle’s velocity at time $t=3$

Answer to Problem 59E

The velocity at $t=3$ is $0\text{units}/\text{sec}$ .

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 v of s is s’ .

Such that

  $v=s\text{'}$

Since

  $s\text{'}=f$

The velocity at $t=3$:

From the graph,

  $v\left(3\right)=s\text{'}\left(3\right)=f\left(3\right)=0$

Thus,

The velocity at $t=3$ is $0\text{units}/\text{sec}$ .

(b)

To determine

Whether the acceleration of particle at time $t=3$ positive or negative.

Answer to Problem 59E

The acceleration of particle is positive at $t=3$ .

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

  $s\text{'}\text{'}\left(3\right)=f\text{'}\left(3\right)$

From the graph,

The slope of f at 3 is positive.

Thus,

  $s\text{'}\text{'}\left(3\right)$ is positive.

Therefore,

The acceleration of particle is positive at $t=3$ .

(c)

To determine

Particle’s position at time $t=3$

Answer to Problem 59E

Particle’s position at 3 seconds is −9 units.

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 triangle, with base of 3 and the height of −6.

Then

The position at 3 seconds:

  $s\left(3\right)={\displaystyle {\int }_0^{3}f\left(x\right)dx=\frac{1}{2}}\times 3\times \left(-6\right)=-9\text{units}$

Therefore,

Particle’s position at 3 seconds is −9 units.

(d)

To determine

Time during which the particle passes through the origin.

Answer to Problem 59E

Particle is at origin when $t=6$ .

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

  

If the particle is at origin,

Then

  $s=0$

Since

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

That means

We need to find some time t where the area between the f and x − axis is 0.

Note that

In the graph, the triangle between 0 and 3 has same area as the triangle between 3 and 6.

For both the triangle,

Base length is 5

And

Height is 6.

Since one triangle is above the x − axis (positive area) and one triangle is below the x − axis (negative axis), the net area is 0 when we add both the triangles.

Thus,

  $s\left(6\right)={\displaystyle {\int }_0^{6}f\left(x\right)dx}=0$

Therefore,

The particle is at origin when $t=6$ .

(e)

To determine

Approximate the zero value of acceleration.

Answer to Problem 59E

The acceleration is zero when $t=7$ .

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

  

If

  $a=s"=0$

Then

The acceleration will be zero.

Thus,

We are required to find the second derivative of s .

Since

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

Then

According to FTC (Fundamental Theorem of Calculus),

We have

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

Then

Second derivative:

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

This implies

If

  $s"=0$

Then

  $f\text{'}\left(t\right)=0$

From the graph,

At $t=7$

The slope of f is 0.

Thus,

We have

  $s\text{'}\text{'}\left(7\right)=f\text{'}\left(7\right)=0$

Therefore,

Acceleration is zero at 7 seconds.

(f)

To determine

Movement of the particle towards and away from the origin.

Answer to Problem 59E

The particle moves away from the origin in the positive direction on the interval [0, 3] and t > 6.

The particle moves towards the origin in the negative direction on the interval [3, 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

  

Note that

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

Such that

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

Thus,

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

And

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

If $s\text{'}>0$ ,

The particle is moving towards the right.

If $s\text{'}<0$ ,

The particle is moving towards the left.

On the interval [0, 3]:

We have

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

Thus,

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

Since it starts at 0 at $t=0$ , and then move towards left.

On the interval [3, 6]:

We have

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

And

  $s\left(3\right)=-9$ [From Part (c) result]

Thus,

The particle moves towards the origin since it was on the negative coordinate axis and moved to the right.

For $t>6$:

Since

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

And

  $s\left(6\right)=0$ [From Part (d) result]

Thus,

The particle moves away from the origin in the positive direction since it was on the origin and then moved to the right.

(g)

To determine

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

Answer to Problem 59E

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 $x=0$ and $x=3$ is smaller than the area above x -axis between $x=3$ and $x=9$ .

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.