# Rearending Drag Racer| University Physics

### PART A. What is $t_{max}$, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?

ANSWER: $t_{max}=\frac{v_0}{a}$

### PART B. Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at $t=t_{max}$), find your distance from the dragster when he started. If you calculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0. Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative).

ANSWER: $D_{start}=\frac{\left(v_0\right)^2}{2a}$

$D_{start}=D_{car}-x_d\left(t_{max}\right)$

$D_{start}=v_0t_{max}-\frac{1}{2}a\left(t_{max}\right)^2$

$D_{start}=v_0\left(\frac{v_0}{a}\right)-\frac{1}{2}a\left(\frac{v_0}{a}\right)^2$

$D_{start}=\frac{\left(v_0\right)^2}{a}-\frac{\left(v_0\right)^2}{2a}$

$D_{start}=\frac{\left(v_0\right)^2}{2a}$

### PART C. Find numerical values for $t_{max}$ and $D_{start}$ in seconds and meters for the (reasonable) values $v_0=60\:mph$ (26.8 m/s) and $a=50\:m/s^2$

ANSWER: $t_{max}=0.54\:s,\:D_{start}=7.2\:m$

$t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s$

$D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m$

The blue curve shows how the car, initially at $x_0$, continues at constant velocity (blue) and just barely touches the accelerating drag car (red) at $t_{max}$.