PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 5, Problem 48P

(a)

To determine

To Calculate: The minimum distance to make the car stop.

(a)

Expert Solution
Check Mark

Answer to Problem 48P

The minimum distance it takes to stop the car is 49.1m

Explanation of Solution

Given Information:

An automobile is going up a 15o grade at a speed of 30 m/s .

The coefficient of static friction between the tires and the road is 0.70 .

Formula Used:

Third equation of motion:

  v12=v02+2as

Where symbols have their usual meanings.

Calculation:

Initial velocity of the automobile v0=30m/s

Final velocity of the automobile v1=0m/s

Let the minimum displacement be smin

Let the acceleration of the car be a

Free body diagram for automobile going up the inclined plane is shown below

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 48P

Use the third kinematics relation

  v12=v02+2asmin(0 m/s)2=v02+2asminsmin=v022a..............(1)

From the free body diagram, along the x-axis

  fs+mgsin15o=maμsFn+mgsin15o=ma...(2)

Along the y-axis:

  Fnmgcos15o=0Fn=mgcos15o...(3)

Substitute (3) in (2)

  μs(mgcos 15o)+mgsin15o=maa=g(μscos 15o+sin 15o)=(9.81  m/s2)(( 0.7)cos 15o+sin 15o)=9.17 m/s2

Substitute the value of acceleration in (1)

  smin=v022a= ( 30 m/s )22( 9.17  m/s 2 )=49.1 m

Conclusion:

Thus, the minimum stopping distance is 49.1 m.

(b)

To determine

To Calculate: The minimum distance it would take to stop if the car were going down the grade.

(b)

Expert Solution
Check Mark

Answer to Problem 48P

  110 m

Explanation of Solution

Given Information:

An automobile is going up a 15o grade at a speed of 30 m/s .

The coefficient of static friction between the tires and the road is 0.70 .

FormulaUsed:

Third equation of motion:

  v12=v02+2as

Where, symbols have their usual meanings

Calculation:

Initial velocity of the car v0=30m/s

Final velocity of the car v1=0m/s

Let the minimum displacement be smin

And let the acceleration of the car be a

Use the third kinematic equation

  v12=v02+2asmin(0 m/s)2=v02+2asminsmin=v022a...(4)

Along the x-axis

  fsmgsin15o=maμsFnmgsin15o=ma...(5)

And along the y-axis we conclude that,

  Fnmgcos15o=0Fn=mgcos15o....(6)

Substitute (6) in (5)

  μs(mgcos 15o)mgsin15o=maa=g(μscos 15osin 15o)=(9.81  m/s2)(( 0.7)cos 15osin 15o)=4.09 m/s2

Substitute the value of acceleration in (1)

  smin=v022a= ( 30 m/s )22( 4 .09 m/s 2 )=110 m

Here, the negative sign indicates that the displacement of the car is in the opposite direction as in part (A) .

Conclusion:

Therefore, the magnitude of displacement is 110 m .

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Chapter 5 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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