Exercise 1 Page 445

Practice makes perfect

a We are asked to solve the given equation for s.

d=s^2/30f To do this we need to use inverse operations. Let's start with multiplying both sides of this equation by 30f.

d=s^2/30f

MultEqn

LHS * 30f=RHS* 30f

30fd=s^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(30fd)=sqrt(s^2)

RootPowToNumber

sqrt(a^n)=a

sqrt(30fd)=s

RearrangeEqn

Rearrange equation

s=sqrt(30fd)

b To find the speed when the skid marks are 120 feet long and the coefficient of fraction is 0.75 we will use the formula we found in Part A. Let's substitute these values for d and f into s=sqrt(30fd) and evaluate.

s=sqrt(30fd)

SubstituteII

d= 120, f= 0.75

s=sqrt(30( 120)( 0.75))

Multiply

s=sqrt(2700)

UseCalc

Use a calculator

s=51.961...

RoundInt

Round to nearest integer

s≈52

When the coefficient of friction for the road is 0.75, the speed of the car is about 52 miles per hour.

c Again, we want to find the speed when the skid marks are 120 feet long. This time, however, the coefficient of the fraction for the road is 1.1. To do this we will use the formula we found in Part A. Let's substitute these values for d and f into s=sqrt(30fd), and evaluate.

s=sqrt(30fd)

SubstituteII

d= 120, f= 1.1

s=sqrt(30( 120)( 1.1))

Multiply

s=sqrt(3960)

UseCalc

Use a calculator

s=62.928...

RoundInt

Round to nearest integer

s≈63