Chapter 2, Problem 28P

(a)

To determine

The stopping distance for a car.

Answer to Problem 28P

The stopping distance for a car is $113\text{ m}$ .

Explanation of Solution

Given:

The initial speed of the car

  $v_0=95\text{ km/h}$

Human reaction time

  $t=1.0\text{ s}$

Acceleration of the car

  $a=-4.0{\text{ m/s}}^2$

Final velocity of the car

  $v=0\text{ m/s}$

Formula used:

The car travels with uniform speed during the reaction time and reaches the position $x_0$ .

Therefore,

  $x_0=v_{0}t$       ……(1)

The brakes are applied when the car is at the position $x_0$ and the car comes to rest after travelling a distance $x$ .

The following equation of motion can be used to determine the position of the car when it comes to rest.

  $v^2=v_0^2+2a\left(x-x_0\right)$       ………………(2)

Calculation:

Express the initial speed of the car in m/s.

  $\begin{array}{c}{v}_0=\left(95\text{ km/h}\right)\left(\frac{1000\text{ m}}{1\text{ km}}\right)\left(\frac{1\text{ h}}{3600\text{ s}}\right)\\ =26.39\text{ m/s}\end{array}$

Using equation (1) calculate the position $x_0$ of the car just before the brakes are applied.

  $\begin{array}{c}{x}_0=v_{0}t\\ =\left(26.39\text{ m/s}\right)\left(1.0\text{ s}\right)\\ =26.39\text{ m}\end{array}$

Rewrite equation (2) for $x$ .

  $x=x_0+\frac{v^2-v_0^2}{2a}$       ……(3)

Substitute the values of the variables in equation (3) and calculate the stopping distance.

  $\begin{array}{c}x=x_0+\frac{v^2-v_0^2}{2a}\\ =\left(26.39\text{ m}\right)+\frac{\left(0\text{ m/s}\right)-\left(26.39\text{ m/s}\right)}{2\left(-4.0{\text{ m/s}}^2\right)}\\ =113\text{ m}\end{array}$

Conclusion:

Thus, the stopping distance for a car for an acceleration of $-4.00{\text{ m/s}}^2$ is $113\text{ m}$ .

(b)

To determine

The stopping distance for a car.

Answer to Problem 28P

The stopping distance for a car is $70\text{ m}$ .

Explanation of Solution

Given:

The initial speed of the car

  $v_0=26.39\text{ m/s}$

Human reaction time

  $t=1.0\text{ s}$

Acceleration of the car

  $a=-8.0{\text{ m/s}}^2$

Final velocity of the car

  $v=0\text{ m/s}$

The position of the car when the brakes are applied will remain same as in (a), since the velocity and the reaction time remain the same.

  $x_0=26.39\text{ m}$

Formula used:

  $x=x_0+\frac{v^2-v_0^2}{2a}$      ……………(3)

Calculation:

Substitute the values of variables in equation (3) and calculate the stopping distance.

  $\begin{array}{c}x=x_0+\frac{v^2-v_0^2}{2a}\\ =\left(26.39\text{ m}\right)+\frac{\left(0\text{ m/s}\right)-\left(26.39\text{ m/s}\right)}{2\left(-8.0{\text{ m/s}}^2\right)}\\ =70\text{ m}\end{array}$

Conclusion:

Thus, the stopping distance for a car for an acceleration of $-8.0{\text{ m/s}}^2$ is $70\text{ m}$ .