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 51P

(a)

To determine

The acceleration of each sprinter.

(a)

Expert Solution
Check Mark

Answer to Problem 51P

The acceleration of Laura and Healan are 5.31m/s2 and 3.74m/s2 respectively.

Explanation of Solution

Section 1:

To determine: The acceleration of the Laura.

Answer: The acceleration of the Laura is 5.31m/s2 .

Given information:

The time taken by Laura and Healan to attain their maximum speeds are 2s and 3s respectively. The time taken to cross the finish line simultaneously by Laura and Healan is 10.4s and the travelled distance for sprinter is 100m

The distance covered by Laura is,

S1=12(t1)2a1+a1(t1)(t1't1) (I)

  • a1 is the acceleration for Laura.
  • S1 is the distance travelled by Laura.
  • t1 is the acceleration time.
  • t1' is the remaining acceleration time for Laura.

Substitute 100m for S1 , 2s for t1 and 8.4s for t1' to find the a1 .

100m=12(2s)2a1+a1(2s)(8.4s)a1=5.31m/s2

Conclusion:

Therefore, the acceleration of Laura is 5.31m/s2 .

Section 2:

To determine: The acceleration of the Healan.

Answer: The acceleration of the Healan is 3.74m/s2 .

Given information:

The time taken by Laura and Healan to attain their maximum speeds are 2s and 3s respectively. The time taken to cross the finish line simultaneously by Laura and Healan is 10.4s .

The distance covered by Healan is,

S2=12(t2)2a2+a2(t2)(t2't2) (II)

  • a2 is the acceleration for Healan.
  • S2 is the distance travelled by the Healan.
  • t2 is the acceleration time.
  • t2' is the remaining acceleration time for Healan.

Substitute 100m for S2 , 3s for t2 and 7.4s for t2' to find the a2 .

100m=12(3s)2a2+a2(3s)(7.4s)a2=3.74m/s2

Conclusion:

Therefore, the acceleration of Healan and 3.74m/s2 .

(b)

To determine

The maximum speeds of Laura and Healan.

(b)

Expert Solution
Check Mark

Answer to Problem 51P

The maximum speeds of Laura and Healan are 10.62m/s and 11.22m/s .

Explanation of Solution

Section 1:

To determine: The maximum speeds of Laura.

Answer: The maximum speeds of Laura is 10.62m/s .

Given information:

The time taken by Laura and Healan to attain their maximum speeds are 2s and 3s respectively. The time taken to cross the finish line simultaneously by Laura and Healan is 10.4s .

Formula to calculate the maximum speed for Laura is,

v1=a1t1

  • v1 is the maximum speed of Laura.

Substitute 5.31m/s2 for a1 and 2s for t1 to find the v1 .

v1=(5.31m/s2)(2s)=10.62m/s

Conclusion:

Therefore, the maximum speed for Laura is 10.62m/s .

Section 2:

To determine: The maximum speeds of Healan.

Answer: The maximum speeds of Healan is 11.22m/s .

Given information:

The time taken by Laura and Healan to attain their maximum speeds are 2s and 3s respectively. The time taken to cross the finish line simultaneously by Laura and Healan is 10.4s .

Formula to calculate the maximum for Healan is,

v2=a2t2

  • v2 is the maximum speed of Healan.

Substitute 3.74m/s2 for a2 and 3s for t2 to find the v2 .

v2=(3.74m/s2)(3s)=11.22m/s

Conclusion:

Therefore, the maximum speed for Healan is 11.22m/s .

(c)

To determine

The sprinter which is ahead at 6.20s from another and also determine the distance by which one sprinter is ahead by another.

(c)

Expert Solution
Check Mark

Answer to Problem 51P

The sprinter is Laura is ahead of Healan by 2.60m .

Explanation of Solution

Given information:

The time taken by Laura and Healan to attain their maximum speeds are 2s and 3s respectively. The time taken to cross the finish line simultaneously by Laura and Healan is 10.4s .

Formula to calculate the distance for Laura at 6s from equation (I) is,

S1=12(t1)2a1+a1(t1)(t1't1)

Formula to calculate the distance for Healan at 6s from equation (II) is,

S2=12(t2)2a2+a2(t2)(t2't2)

The difference of distance travelled by two sprinters is,

ΔS=S1S2

Substitute 12(t1)2a1+a1(t1)(t1't1) for S1 and 12(t2)2a2+a2(t2)(t2't2) for S2 in the above equation.

ΔS=(12(t1)2a1+a1(t1)(t1't1))(12(t2)2a2+a2(t2)(t2't2))

Substitute 5.31m/s2 for a1 , 6.20s for t1' , 2s for t1 , 3.74m/s2 for a2 , 6s for t2' and 3s for t2 to find ΔS .

ΔS=[{12(2s)25.31m/s2+5.31m/s2(2s)(6s2s)}{12(3s)23.74m/s2+3.74m/s2(3s)(6s3s)}]=53.1m50.5m=2.6m

The positive sign shows that Laura is ahead of Healan.

Conclusion:

Therefore, the sprinter Laura is ahead of Healan by 2.60m .

(d)

To determine

The maximum distance by which Healan is behind Laura and time at which maximum distance occurs.

(d)

Expert Solution
Check Mark

Answer to Problem 51P

The maximum distance by which Healan is behind Laura is 4.46m at time 2.84s .

Explanation of Solution

Given information:

The time taken by Laura and Healan to attain their maximum speeds are 2s and 3s respectively. The time taken to cross the finish line simultaneously by Laura and Healan is 10.4s .

Maximum distance between runners occurs when each has the same velocity setting the equal to each other.

Laura has already reached her maximum speed while Healan is still accelerating so,

10.62m/s=(3.74m/s2)tt=2.84s

Formula to calculate the distance for Laura at 2.84s from equation (I) is,

S1=12(t1)2a1+a1(t1)(tt1)

  • t is the time at which maximum distance occurs.

Formula to calculate the distance for Healan from equation (II) is,

S2=12(t)2a2

The maximum distance by which Healan is behind Laura is,

ΔS=S1S2

Substitute 12(t1)2a1+a1(t1)(tt1) for S1 and 12(t)2a2 for S2 in the above equation.

ΔS=(12(t1)2a1+a1(t1)(tt1))12(t)2a2

Substitute 5.31m/s2 for a1 , 2.84s for t1' , 2s for t1 , 5.31m/s2 for a2 and 2.84s for t to find ΔS .

ΔS={12(2s)25.31m/s2+5.31m/s2(2s)(2.84s2s)}12(2.84s)23.74m/s2=19.54m15.08m=4.46m

Conclusion:

Therefore, the maximum distance by which Healan is behind Laura is 4.46m at time 2.84s .

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Chapter 2 Solutions

Principles of Physics: A Calculus-Based Text

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Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY