Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 2, Problem 111P

(a)

To determine

The velocity as the function of time for the given time interval.

(a)

Expert Solution
Check Mark

Answer to Problem 111P

The velocity as the function of time interval is v(t)=(0.10m/s3)t2+9.5m/s .

Explanation of Solution

Given:

The acceleration of the particle is given by ax=(0.20m/s3)t .

The velocity of particle at t=0 is 9.5m/s .

The position of particle at t=0 is 5.0m .

Formula used:

Write expression for the acceleration of the particle.

  dv(t)dt=a(t)

Here, a is the acceleration of the particle and dv(t)dt is the rate of change of velocity.

  dv(t)=a(t)dt

Substitute (0.20m/s3) for a(t) in above expression and integrate.

  dv(t)=(0.20m/ s 3 )tdt

Simplify above expression.

  v(t)=(0.20m/ s 3)2t2+C............. (1)

Calculation:

Substitute 0 for t in equation (1).

  v(0)=( 0.20m/ s 3 )2(0)2+CC=v(0)

Substitute 9.5m/s for v(0) in above expression.

  C=9.5m/s

Substitute 9.5m/s for C in equation (1).

  v(t)=( 0.20m/ s 3 )2t2+9.5m/s=(0.10m/ s 3)t2+9.5m/s

Conclusion:

Thus, the velocity as the function of time interval is v(t)=(0.10m/s3)t2+9.5m/s .

(b)

To determine

The position as function of time for the given time interval.

(b)

Expert Solution
Check Mark

Answer to Problem 111P

The position as function of time for the given interval is x(t)=(0.10m/ s 3)t33+(9.5m/s)t5.0m .

Explanation of Solution

Given:

The acceleration of the particle is given by ax=(0.20m/s3)t .

The velocity of particle at t=0 is 9.5m/s .

The position of particle at t=0 is 5.0m .

Formula used:

Write expression for the velocity of the particle as function of time for given time period.

  v(t)=(0.10m/s3)t2+9.5m/s

Write expression for velocity of the particle.

  dx(t)dt=v(t)

Here, v is the velocity of the particle and dxdt is the rate of change of position of the particle.

Rearrange above expression for dx .

  dx(t)=v(t)dt  ........(2)

Calculation:

Substitute (0.10m/s3)t2+9.5m/s for v(t) in equation (2) and integrate.

  dx(t)=[( 0.10m/ s 3 )t2+9.5m/s]dt

Simplify above expression.

  x(t)=(0.10m/ s 3)t33+(9.5m/s)t+D

  ........(3)

Substitute 0 for t in above expression.

  x(0)=( 0.10m/ s 3 ) ( 0 )33+(9.5m/s)(0)+Dx(0)=D

Substitute 5m for x(0) in above expression.

  D=5.0m

Substitute 5.0m for D in equation (3).

  x(t)=( 0.10m/ s 3 )t33+(9.5m/s)t+(5.0m)x(t)=( 0.10m/ s 3 )t33+(9.5m/s)t5.0m

Conclusion:

Thus, the position as function of time for the given interval is x(t)=(0.10m/ s 3)t33+(9.5m/s)t5.0m .

(c)

To determine

The average velocity for the given time interval and compare to the average of instantaneous velocities of starting and ending times.

(c)

Expert Solution
Check Mark

Answer to Problem 111P

The average velocity for the given time interval is 12.8m/s and it is not equal to the average of instantaneous velocities of the start and ending times.

Explanation of Solution

Given:

The acceleration of the particle is given by ax=(0.20m/s3)t .

The velocity of particle at t=0 is 9.5m/s .

The position of particle at t=0 is 5.0m .

Formula used:

Write expression for average velocity of the particle.

  vav=1Δtt=t1t2v(t)dt............. (4)

Write expression for instantaneous velocity of the particle.

  v(t)=(0.10m/s3)t2+9.5m/s

  ........(5)

Write expression for average of instantaneous velocities for t=0 and t=10 .

  vav=v(10)+v(0)2............. (6)

Calculation:

Substitute 0s for t1 , 10s for t2 , 10s for Δt and (0.10m/s3)t2+9.5m/s for v(t) in equation (4).

  vav=110t=010[( 0.10m/ s 3 ) t 2+9.5m/s]dtvav=110( ( 0.10m/ s 3 ) t 3 3+( 9.5m/s )t)010vav=110[( 0.10m/ s 3 ) ( 10 )33+(9.5m/s)(10)]vav=12.8m/s

Substitute 0s for t in equation (5).

  v(0)=(0.10m/ s 3)(0)2+9.5m/sv(0)=9.5m/s

Substitute 10s for t in equation (5).

  v(10)=(0.10m/ s 3)(10s)2+9.5m/sv(10)=[10+9.5]m/sv(10)=19.5m/s

Substitute 19.5m/s for v(10) and 0m/s for v(0) in equation (6).

  vav=19.5m/s+9.5m/s2vav=14.5m/s

Conclusion:

Thus, the average velocity for the given time interval is 12.8m/s and it is not equal to the average of instantaneous velocities of the start and ending times.

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Chapter 2 Solutions

Physics for Scientists and Engineers

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