Concept explainers
(a)
Particle’s velocity at time
(a)
Answer to Problem 59E
The velocity at
Explanation of Solution
Given information:
Position at time t (sec) of a particle moving along a coordinate axis:
Where,
f is the differentiable function
The velocity v of s is s’ .
Such that
Since
The velocity at
From the graph,
Thus,
The velocity at
(b)
Whether the acceleration of particle at time
(b)
Answer to Problem 59E
The acceleration of particle is positive at
Explanation of Solution
Given information:
Position at time t (sec) of a particle moving along a coordinate axis:
Where,
f is the differentiable function
The acceleration of s is s”.
Since
Then
According to FTC (Fundamental Theorem of Calculus),
Second derivative:
That means
From the graph,
The slope of f at 3 is positive.
Thus,
Therefore,
The acceleration of particle is positive at
(c)
Particle’s position at time
(c)
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:
Where,
f is the differentiable function
The region between
Then
The position at 3 seconds:
Therefore,
Particle’s position at 3 seconds is −9 units.
(d)
Time during which the particle passes through the origin.
(d)
Answer to Problem 59E
Particle is at origin when
Explanation of Solution
Given information:
Position at time t (sec) of a particle moving along a coordinate axis:
Where,
f is the differentiable function
If the particle is at origin,
Then
Since
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,
Therefore,
The particle is at origin when
(e)
Approximate the zero value of acceleration.
(e)
Answer to Problem 59E
The acceleration is zero when
Explanation of Solution
Given information:
Position at time t (sec) of a particle moving along a coordinate axis:
Where,
f is the differentiable function
If
Then
The acceleration will be zero.
Thus,
We are required to find the second derivative of s .
Since
Then
According to FTC (Fundamental Theorem of Calculus),
We have
Then
Second derivative:
This implies
If
Then
From the graph,
At
The slope of f is 0.
Thus,
We have
Therefore,
Acceleration is zero at 7 seconds.
(f)
Movement of the particle towards and away from the origin.
(f)
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:
Where,
f is the differentiable function
Note that
Such that
Thus,
The particle is at origin at
And
If
The particle is moving towards the right.
If
The particle is moving towards the left.
On the interval [0, 3]:
We have
Thus,
The particle moves away from the origin in the negative direction.
Since it starts at 0 at
On the interval [3, 6]:
We have
And
Thus,
The particle moves towards the origin since it was on the negative coordinate axis and moved to the right.
For
Since
And
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)
Side of the origin for the particle at time
(g)
Answer to Problem 59E
The particle lies on positive side of the origin at time
Explanation of Solution
Given information:
Position at time t (sec) of a particle moving along a coordinate axis:
Where,
f is the differentiable function
Since the particle starts at the origin,
Then
Also,
The area below the x -axis between
Thus,
Therefore,
The particle lies on the positive side of the origin at 9 seconds.
Chapter 6 Solutions
Calculus: Graphical, Numerical, Algebraic: Solutions Manual
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