BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.11, Problem 36E
To determine

To find: The maximum speed (s) of the car when it stop in 200ft .

Expert Solution

Answer to Problem 36E

The maximum speed of the car is 46mi/h .

Explanation of Solution

Given:

The stopping distance (D) is directly proportional to square of speed (s) .

Speed of the car (s) is 40mi/h .

Stopping distance (D) for speed of 40mi/h is 150ft .

Stopping distance (D) for which maximum speed is calculated is 200ft .

Calculation:

Direct variation:

If the quantities a and b are related by the equation a=Cb then for C0 , a is directly proportional to b and C is called constant of proportionality.

The given variation is,

D=k(s2) (1)

First calculate constant of proportionality k .

Substitute 200 for D and 40 for s in equation (1) and solve for k .

D=k(s2)150=k(402)150=k(1600)

Multiply both the sides by (11600) .

(11600)150=(11600)k(1600)1501600=k0.0937=k0.0937=k (1)

Now, calculate speed for stopping distance of 200ft .

Substitute 0.0937 for k , 200 for D in equation (1) and solve for s .

D=k(s2)200=0.0937(s2)2000.0937=s22134.47=s2

Take square root of above equation.

s=46.2

Thus, the value of maximum speed for stopping distance of 200ft is approximately 46mi/h .

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