Rearending Drag Racer| University Physics

Problem:

To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is “burning out” at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

You drive at a constant speed of v_0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a. Let the time at which the dragster starts to accelerate be t=0.

The figure shows a car moving to the right with speed v 0 toward a stopped dragster, which stands in front of the lights facing to the right.

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}.

Two graphs for the position as a function of time are shown on the same set of axes. Time is measured from to 0.6 seconds on the x-axis. The position is measured from – 9 to 9 meters on the y-axis. The distance on the blue graph increases linearly from 0 seconds and x 0, which is between - 9 and - 6 meters, to 0.6 seconds and 9 meters. The distance on the red graph smoothly increases from 0 seconds and 0 meters to 0.6 seconds and 9 meters, forming a convex curve. Time t max is marked on the x-axis between 0.5 and 0.55 seconds.