Exercise 1 Page 445

Practice makes perfect

a We are asked to solve the given equation for

s .

d = \frac{s ^{2}}{3 0 f}

To do this we need to use inverse operations. Let's start with multiplying both sides of this equation by

30 f .

d = \frac{s ^{2}}{3 0 f}

MultEqn

$LHS \cdot 30 f = RHS \cdot 30 f$

30 f d = s^{2}

SqrtEqn

$LHS = RHS$

30 f d = s^{2}

RootPowToNumber

$n a^{n} = a$

30 f d = s

RearrangeEqn

Rearrange equation

s = 30 f d

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 = 30 f d

and evaluate.

s = 30 f d

SubstituteII

$d = 120$ , $f = 0.75$

s = 30 (120) (0.75)

Multiply

s = 2700

UseCalc

Use a calculator

s = 51.961 \dots

RoundInt

Round to nearest integer

s \approx 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 = 30 f d,

and evaluate.

s = 30 f d

SubstituteII

$d = 120$ , $f = 1.1$

s = 30 (120) (1.1)

Multiply

s = 3960

UseCalc

Use a calculator

s = 62.928 \dots

RoundInt

Round to nearest integer

s \approx 63

When the coefficient of friction for the road is

1.1,

the speed of the car is about

63

miles per hour.

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