Chapter 3.3, Problem 95AYU

Stopping Distance 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?

To determine

To calculate:

a. The distance feet will a car travelling with a speed of $45mph$ take to stop on a dry level concrete.

b. The maximum speed we can travel in order to avoid an accident occurring $200feet$ ahead.

c. Representation of $1.1v$.

Answer to Problem 95AYU

a. $171feet$

b. $49.2mph$

c. $1.1v$ might represent the reaction time.

Explanation of Solution

Given:

The relation between the stopping distance, $d$ and the speed of the car, $v$ is

$d=1.1v+0.06v^2$

Formula used:

The quadratic equation for finding the value of $x$ in $f(x)=a{x}^2+bx+c,a\ne 0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

a. When $v=45mph$, the stopping distance is

$d=1.1(45)+0.06{(45)}^2=171$

Therefore, the stopping distance is $171feet$.

b. When $200feet$, the maximum speed that we can travel is

$200=1.1v+0.06v^2$

$\Rightarrow 0.06v^2+1.1v-200$

$\Rightarrow v=\frac{-1.1\pm \sqrt{{1.1}^2-4(0.06) (-200)}}{2(0.06)}$

$=\frac{-1.1\pm \sqrt{1.21+48}}{0.12}$

$=\frac{-1.1\pm 7.015}{0.12}=49.2,-67.63$

Here, the negative value can be excluded as the speed can never be negative.

Therefore, the maximum speed when $d=200feet$ is $49.2mph$.

c. The stopping distance is the distance travelled at a particular speed in a particular time of reacting. Therefore, $1.1v$ might represent the reaction time.