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A car weighs 3000 lb m and is travelling 60 mph when it has to make an emergency stop. The car comes to a stop 5 seconds after the brakes are applied. A. Assuming the rate of deceleration is constant, what force is required? B. Assuming the rate of deceleration is constant, how much distance is covered before the car comes to a stop?

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Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
Publisher: Cengage Learning
ISBN: 9781111580704

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Chapter
Section
BuyFindarrow_forward

Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
Publisher: Cengage Learning
ISBN: 9781111580704
Chapter 1.7, Problem 18P
Textbook Problem
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A car weighs 3000 lbm and is travelling 60 mph when it has to make an emergency stop. The car comes to a stop 5 seconds after the brakes are applied.

A. Assuming the rate of deceleration is constant, what force is required?

B. Assuming the rate of deceleration is constant, how much distance is covered before the car comes to a stop?

A)

Interpretation Introduction

Interpretation:

The amount of force required to stop the car when the deceleration is constant.

Concept Introduction:

The acceleration of the car.

a=vfvit

Here, final velocity is vf, initial velocity is vi, and time is t.

The required force to stop the car:

F=M×a

Here, mass is M and acceleration is a.

Explanation of Solution

Given information:

Weight of the car =3000lbm

Speed of the car =60mph

Time taken to stop the car =5s

Calculate the acceleration of the car.

a=vfvit        (1)

Here, final velocity is vf, initial velocity is vi, and time is t.

Substitute 0 for vi, 60 mph for vf, and 5 s for t in equation (1).

a=60mph05s=60mph×(1.466fts)5s=17.592ft/s2

Calculate the force required to stop the car

B)

Interpretation Introduction

Interpretation:

The distance covered by the car before it comes to stop.

Concept Introduction:

The distance covered before the car comes to a stop.

df=a×t22+vf×t+di

Here, initial distance is di, final distance is df, and initial velocity is vi.

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