Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 2, Problem 13P

A particle starts from rest and accelerates as shown in Figure P2.13. Determine (a) the particle’s speed at t = 10.0 s and at t = 20.0 s, and (b) the distance traveled in the first 20.0 s.

Chapter 2, Problem 13P, A particle starts from rest and accelerates as shown in Figure P2.13. Determine (a) the particles

(a)

Expert Solution
Check Mark
To determine

The particle’s speed at t=10.0 s and at t=20.0 s.

Answer to Problem 13P

The particle’s speed at t=10.0 s is 20 m/s and at t=20.0 s is 5 m/s.

Explanation of Solution

Write the equation for the final velocity of a particle.

  v=v0+Δv        (I)

Here, v is the final velocity of a particle, v0 is the initial velocity of a particle and Δv is the change in velocity.

The area under acceleration versus time graph under specific time intervals gives the change in velocity of the particle during the time interval. Speed is the magnitude of velocity.

The sides of a unit rectangle in the graph are 1 m/s2 and 5 s so that the area of a unit rectangle will be (1 m/s2)(5 s)=5 m/s .

The area under the graph in the interval 0<t<10.0 s has four unit rectangles so that the area under the curve in the time interval is 4(5 m/s)=20 m/s .

Since the particle starts from rest, its velocity at t=0 s will be 0 m/s.

Substitute 0 m/s for v0 and 20 m/s for Δv in equation (I) to find v at t=10.0 s.

  v=0 m/s+20 m/s=20 m/s

The area under the graph in the interval 10.0<t<20.0 s has three unit rectangles in the negative direction so that the area under the curve in the time interval is 3(5 m/s)=15 m/s.

Substitute 20 m/s for v0 and 15 m/s for Δv in equation (I) to find v at t=20.0 s.

  v=20 m/s+(15 m/s)=5 m/s

Conclusion:

Therefore, the particle’s speed at t=10.0 s is 20 m/s and that at t=20.0 s is 5 m/s.

(b)

Expert Solution
Check Mark
To determine

The distance travelled in the first 20.0 s.

Answer to Problem 13P

The distance travelled in the first 20.0 s is 263 m.

Explanation of Solution

The area under velocity versus time graph under specific time intervals gives the displacements during the time interval.

In part (a), it is found that the velocity at t=0 s is 0 m/s, at t=10.0 s is 20 m/s and that at t=20.0 s is 5 m/s . In the given acceleration versus time graph, the acceleration is zero in the interval 10.0 s<t<15.0 s . This implies there is no velocity change during the interval or the velocity at t=15.0 s is same as velocity at t=10.0 s . These data can be used to plot the velocity versus time graph for the given time interval.

The velocity versus time graph is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 2, Problem 13P

The area from 010 s is that of a triangle.

Write the equation for the area of a triangle.

  At=12bh        (II)

Here, At is the area of the triangle. b is the base and h is the height of the triangle.

In figure 1, the base of the triangle from 010 s is 10 s and the height of the triangle is 20 m/s .

Substitute 10 s for b and 20 m/s for h in equation (II) to find At .

  At010=12(10 s)(20 m/s)=100 m

Here, At010 is the area under the graph from 010 s .

The area from 1015 s is that of a rectangle.

Write the equation for the area of a rectangle.

  Ar=ld        (III)

Here, Ar is the area of the rectangle. l is the length of the rectangle and d is the breadth of the rectangle.

In figure 1, the length of the rectangle from 1015 s is 5 s and the breadth of the rectangle is 20 m/s .

Substitute 5 s for l and 20 m/s for d in equation (III) to find Ar .

  Ar1015=(5 s)(20 m/s)=100 m

Here, Ar1015 is the area under the graph from 1015 s .

The area from 1520 s is the sum of the area of a triangle and a rectangle.

In figure 1, the base of the triangle from 1520 s is 5 s and the height of the triangle is 20 m/s5 m/s=15 m/s .

Substitute 5 s for b and 15 m/s for h in equation (II) to find At .

  At1520=12(5 s)(15 m/s)=37.5 m

Here, At1520 is the area of the triangle under the graph from 1520 s .

In figure 1, the length of the rectangle from 1520 s is 5 s and the breadth of the rectangle is 5 m/s .

Substitute 5 s for l and 5 m/s for d in equation (III) to find Ar .

  Ar1520=(5 s)(5 m/s)=25 m

Here, Ar1520 is the area of the rectangle under the graph from 1520 s .

Write the equation for total distance travelled.

  Δx=At010+Ar1015+At1520+Ar1520        (IV)

Here, Δx is the total distance travelled.

Conclusion:

Substitute 100 m for At010, 100 m for Ar1015, 37.5 m for At1520 and 25 m for Ar1520 in equation (IV) to find Δx .

  Δx=100 m+100 m+37.5 m+25 m=263 m

Therefore, the distance travelled in the first 20.0 s is 263 m.

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Chapter 2 Solutions

Principles of Physics: A Calculus-Based Text

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