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

#### PART A. What is $\displaystyle t_{max}$ $\displaystyle t_{max}$, 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? $\displaystyle t_{max}=\frac{v_0}{a}$

#### 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 $\displaystyle t=t_{max}$ $\displaystyle t=t_{max}$), 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). $\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}$ $\displaystyle D_{start}=D_{car}-x_d\left(t_{max}\right)$ $\displaystyle D_{start}=v_0t_{max}-\frac{1}{2}a\left(t_{max}\right)^2$ $\displaystyle D_{start}=v_0\left(\frac{v_0}{a}\right)-\frac{1}{2}a\left(\frac{v_0}{a}\right)^2$ $\displaystyle D_{start}=\frac{\left(v_0\right)^2}{a}-\frac{\left(v_0\right)^2}{2a}$ $\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}$

#### PART C. Find numerical values for $\displaystyle t_{max}$ $\displaystyle t_{max}$ and $\displaystyle D_{start}$ $\displaystyle D_{start}$ in seconds and meters for the (reasonable) values $\displaystyle v_0=60\:mph$ $\displaystyle v_0=60\:mph$ (26.8 m/s) and $\displaystyle a=50\:m/s^2$ $\displaystyle a=50\:m/s^2$ $\displaystyle t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s$ $\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m$
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 $\displaystyle t_{max}$.