Chapter 2, Problem 29P

To determine

To show: The equation for the stopping distance of a car is $d_{\text{S}}=v_0{t}_{\text{R}}-\frac{v_0^2}{2a}$ , where $v_0$ is the initial speed of the car, $t_{\text{R}}$ is the driver reaction time and $a$ is the constant negative acceleration.

Explanation of Solution

Given:

A car moving with a speed $v_0$ would travel a distance $x_0$ during the driver reaction time. The brakes are applied when the car is at $x_0$ .The car suffers negative acceleration and comes to rest.

Formula used:

The position of the car when the brakes are applied is given by the expression,

  $x_0=v_0{t}_{\text{R}}$        ……(1)

Write the equation of motion for the constant decoration of the car.

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

Rewrite the equation (2) for $x$ , the position of the car when it comes to rest.

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

The position of the car when it comes to rest gives the stopping distance. Hence,

  $d_{\text{S}}=x$      ……………(4)

Calculation:

Substitute 0 for v , and equations (1) and (4) in equation (3) .

  $d_{\text{S}}=v_0{t}_{\text{R}}+\frac{0-v_0^2}{2a}$

Therefore,

  $d_{\text{S}}=v_0{t}_{\text{R}}-\frac{v_0^2}{2a}$

Conclusion:

Thus, it can be proved that the equation for the stopping distance of a car is $d_{\text{S}}=v_0{t}_{\text{R}}-\frac{v_0^2}{2a}$ , where $v_0$ is the initial speed of the car, $t_{\text{R}}$ is the driver reaction time and $a$ is the constant negative acceleration.