To demonstrate the tremendous acceleration of a top fuel Drag Racer| University Physics


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 \displaystyle 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, \displaystyle a. Let the time at which the dragster starts to accelerate be \displaystyle 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 \displaystyle 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:

\displaystyle 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 \displaystyle 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:

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

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

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

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

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

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


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


Answer:

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

\displaystyle 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 \displaystyle x_0, continues at constant velocity (blue) and just barely touches the accelerating drag car (red) at \displaystyle t_{max}.


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