Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 2, Problem 46PQ

In Example 2.6, we considered a simple model for a rocket launched from the surface of the Earth. A better expression for the rocket’s position measured from the center of the Earth is given by

y ( t ) = ( R 3 / 2 + 3 g 2 R t ) 2 / 3 j ^

where R is the radius of the Earth (6.38 × 106 m) and g is the constant acceleration of an object in free fall near the Earth’s surface (9.81 m/s2).

a. Derive expressions for v y ( t ) and a y ( t ) .

b. Plot y(t), vy(t), and ay(t). (A spreadsheet program would be helpful.)

c. When will the rocket be at y = 4 R ?

d. What are v y and a y when y = 4 R ?

(a)

Expert Solution
Check Mark
To determine

The expressions for vy(t) and ay(t).

Answer to Problem 46PQ

The expression for vy(t) is vy(t)=2g2R(R3/2+3g2Rt)1/3j^ and the expression for ay(t) is ay(t)=(gR2)(R3/2+3g2Rt)4/3j^.

Explanation of Solution

Write the given expression for the position vector.

  yy(t)=(R3/2+3g2Rt)2/3j^                                                                                  (I)

Here, y(t) is the position vector, R is the radius of the Earth, g is the acceleration due to gravity, and t is the time.

Velocity is the time derivative of position vector. Write the equation for velocity.

  vy(t)=dy(t)dt                                                                                                   (II)

Here, vy(t) is the velocity.

Acceleration is the time derivative of velocity. Write the expression for acceleration.

  ay(t)=dvy(t)dt                                                                                                (III)

Here, ay(t) is the acceleration.

Conclusion:

Put equation (I) in equation (II).

  vy(t)=d[(R3/2+3g2Rt)2/3j^]dt=23(R3/2+3g2Rt)1/33g2Rj^=2g2R(R3/2+3g2Rt)1/3j^                                                               (IV)

Put equation (IV) in equation (III).

  ay(t)=d[2g2R(R3/2+3g2Rt)1/3j^]dt=2g2R(13)(R3/2+3g2Rt)4/33g2Rj^=(gR2)(R3/2+3g2Rt)4/3j^                                              (V)

Therefore, the expression for vy(t) is vy(t)=2g2R(R3/2+3g2Rt)1/3j^ and the expression for ay(t) is ay(t)=(gR2)(R3/2+3g2Rt)4/3j^ .

(b)

Expert Solution
Check Mark
To determine

Plots of y(t), vy(t) and ay(t) .

Answer to Problem 46PQ

The plot of y(t) is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 2, Problem 46PQ , additional homework tip  1

The plot of vy(t) is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 2, Problem 46PQ , additional homework tip  2

And the plot of ay(t) is

Physics for Scientists and Engineers: Foundations and Connections, Chapter 2, Problem 46PQ , additional homework tip  3

Explanation of Solution

The graph of position versus time of an object gives the position of the object at different instant of time. The slope of the position versus time graph gives the magnitude of the velocity of the object. In velocity versus time graph of an object, its velocity at different instants of time is plotted. The slope of this graph gives the magnitude of acceleration of the object. In acceleration versus time graph, acceleration is plotted as a function of time.

The plot of y(t) is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 2, Problem 46PQ , additional homework tip  4

The plot of vy(t) is shown in figure 2.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 2, Problem 46PQ , additional homework tip  5

From the figure it is clear that the rocket has maximum velocity when it starts its motion and the velocity decreases with time. The graph has negative slope implying the acceleration is negative.

The plot of ay(t) is shown in figure 3.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 2, Problem 46PQ , additional homework tip  6

From the figure, it is clear that the rocket has negative acceleration.

(c)

Expert Solution
Check Mark
To determine

The time at which the rocket will be at y=4R .

Answer to Problem 46PQ

The time at which the rocket will be at y=4R is 2661 s.

Explanation of Solution

Equation (I) can be used to determine the time at which the rocket will be at y=4R .

Substitute 4R for y in equation (I).

  4R=(R3/2+3g2Rt)2/3

Take the power (3/2) of the above equation and rewrite it for t .

  (4R)3/2=R3/2+3g2Rtt=R3/2(43/21)3g2R=(43/21)32Rg                                                                                    (VI)

Conclusion:

Given that the radius of the Earth is 6.38×106 m and the acceleration due to gravity is 9.81 m/s2 .

Substitute 6.38×106 m for R and 9.81 m/s2 for g in equation (VI) to find t .

  t=(43/21)32(6.38×106 m)9.81 m/s2=2661 s

Therefore, the time at which the rocket will be at y=4R is 2661 s.

(d)

Expert Solution
Check Mark
To determine

The value of vy and ay when y=4R .

Answer to Problem 46PQ

The value of vy when y=4R is (5590 m/s)j^ and the value of ay is (0.613 m/s2)j^ .

Explanation of Solution

Equation (IV) can be used to determine the value of vy and equation (V) can be used to determine the value of ay .

Conclusion:

Substitute 2661 s for t, 6.38×106 m for R and 9.81 m/s2 for g in equation (IV) to find vy .

  vy=29.81 m/s22(6.38×106 m)((6.38×106 m)3/2+39.81 m/s22(6.38×106 m)(2661 s))1/3j^=(5590 m/s)j^

Substitute 2661 s for t, 6.38×106 m for R and 9.81 m/s2 for g in equation (V) to find ay .

  ay(t)=((9.81 m/s2)(6.38×106 m)2)((6.38×106 m)3/2+3(9.81 m/s2)2(6.38×106 m)(2661 s))4/3j^=(0.613 m/s2)j^

Therefore, the value of vy when y=4R is (5590 m/s)j^ and the value of ay is (0.613 m/s2)j^ .

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Chapter 2 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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