Chapter 6.3, Problem 40PPS

(a)

To determine

Find the speed of car that skids 25 feet.

Answer to Problem 40PPS

The speed of the car is about 21 miles per hour.

Explanation of Solution

Given:

After an accident, police can determine how fast a car was travelling before the driver put on his or her brakes by using the equation $v=\sqrt{30fd}$ .In this equation, v represents the speed in miles per hour, f represents the coefficient of friction, and d represents the length of the skid marks in feet.Assume that

f = 0.6

Calculation:

Given:

  $v=\sqrt{30fd}$

Substitute :

d = 25 feet

f = 0.6

  $\begin{array}{l}v=\sqrt{30fd}\\ \Rightarrow v=\sqrt{30\times 0.6\times 25}\\ \Rightarrow v=\sqrt{30\times 5^2\times \frac{6}{10}}\\ \Rightarrow v=5\sqrt{18}\\ \Rightarrow v=5\sqrt{3^2\cdot 2}\\ \Rightarrow v=15\sqrt{2}\approx 21\end{array}$

So, the speed of the car is about 21 miles per hour.

(b)

To determine

If the car is going 35 miles per hour, find the distance at which it would take to stop.

Answer to Problem 40PPS

The car would stop after about 68.1 feet.

Explanation of Solution

Given:

After an accident, police can determine how fast a car was travelling before the driver put on his or her brakes by using the equation $v=\sqrt{30fd}$ .In this equation, v represents the speed in miles per hour, f represents the coefficient of friction, and d represents the length of the skid marks in feet.Assume that

f = 0.6

Calculation:

Given:

  $v=\sqrt{30fd}$

Substitute :

v = 35 miles per hour

f = 0.6

Solve for d:

  $\begin{array}{l}v=\sqrt{30fd}\\ \Rightarrow 35=\sqrt{30\times 0.6\times d}\\ \Rightarrow {35}^2=18d\mathrm{.................}[\text{square both sides]}\\ \Rightarrow d=\frac{1225}{18}\mathrm{.................}[\text{divide each side by 18]}\\ \Rightarrow d=68.05555556\approx 68.1\end{array}$

So, the car would stop after about 68.1 feet.

(c)

To determine

If the speed of a car is doubled , explain if the skid be twice long or not.

Answer to Problem 40PPS

No

Explanation of Solution

Given:

After an accident, police can determine how fast a car was travelling before the driver put on his or her brakes by using the equation $v=\sqrt{30fd}$ .In this equation, v represents the speed in miles per hour, f represents the coefficient of friction, and d represents the length of the skid marks in feet.Assume that

f = 0.6

Calculation:

Given:

  $v=\sqrt{30fd}$

Substitute :

f = 0.6

Solve for d:

  $\begin{array}{l}v=\sqrt{30fd}\\ \Rightarrow v=\sqrt{30\times 0.6\times d}\\ \Rightarrow {\left(v\right)}^2=18d\mathrm{...................}[\text{square both sides]}\\ \Rightarrow v^2=18d\\ \Rightarrow d=\frac{v^2}{18}\mathrm{......................}[\text{divide each side by 18]}\end{array}$

If the velocity is doubled, the skid will be 4 times.

  $\begin{array}{l}d=\frac{v^2}{18}\\ If\text{ v is doubled:}\\ {\text{v}}_1=2v\\ d_1=\frac{{\left(v_1\right)}^2}{18}\\ d_1=\frac{{\left(2v\right)}^2}{18}\\ d_1=4\left(\frac{v^2}{18}\right)\\ d_1=4d\end{array}$

Hence , the skid is not doubled, if the velocity is doubled.