Q. 93 - Precalculus Enhanced with Graphing Utilities

Question

An accepted relationship between

stopping distance, d (in feet), and the speed of a car, v (in mph), is $d = 1.1v + 0.06v^2$ on dry, level concrete.

(a) How many feet will it take a car traveling 45 mph to stop on dry, level concrete?

(b) If an accident occurs 200 feet ahead of you, what is the maximum speed you can be traveling to avoid being involved?

(c) What might the term $1.1v$ represent?

Step-by-Step Solution

Answer

a)it will take 171 feet for a car travelling 45mph

b)The speed of the car should be less than 49mph

c)1.1v represents a values which is greater 10% greater than v

Part a) Step 1: Given information

We are given a equation $d = 1.1v + 0.06v^2$

Part a) Step 2: To find the feet substitute v = 45 in the equation we get

We have,

$$\begin{gathered} d = 1.1v + 0.06v^2 \\ d=1.1\left(45\right)+0.06{\left(45\right)}^2 \\ d=171 \end{gathered}$$

Part b) Step 1: Put d = 200 and solve theinequality

We get

$$\begin{gathered} 1.1v+0.06v^2<200 \\ 0.06v^2+1.1v-200<0 \end{gathered}$$

On further solving We get

$v=49 ot v=-68$

As speed cannot be negative $v=49$

Part c) Step 4) Explanation

The term $1.1v$ represents a value who is 10% greater than v

Step 5: Conclusion

a)it will take 171 feet for a car travelling 45mph

b)The speed of the car should be less than 49mph

c)1.1v represents a values which is greater 10% greater than v