Question

A dragster travels 1/4 mi in 6.7 s. Assuming that acceleration is constant and the dragster is initially at rest, what is its velocity when it crosses the finish line?

Please show me step by step

Answer 1

I'd say it'll be the same as the one he has there, since we're assuming is constant and never changed

Answer 2

Position for a constant acceleration is given by the following\[s(t)=\frac{at^2}{2}+v_0t+s_0\]In this case, vsub0 and ssub0 are both zero, so we have\[s(t)=\frac{at^2}{2}\]When combined with our given data,\[s(6.7)=1320=\frac{a(6.7)^2}{2} \implies \frac{2640}{6.7^2} \approx 58.8 \frac{ft}{s^2}=a\]This gives us our position funtion\[s(t)= \frac{58.8t^2}{2} \implies s'(t)=v(t)=58.8t \implies v(6.7) \approx 394 \frac{ft}{s}\]If you like miles per hour better, the conversion is 60 mph/88 fps. It gives\[394 \frac{ft}{s} \approx 269~mph\]I rounded a couple of times, so you might get a slightly different answer by carrying all the numbers through the calculations to the end, and only round off once.

Answer 3

By the way, 269 mph is really getting with the program for a car that runs the quarter in 6.70. This combination of speed and time indicates a lot of horsepower, but a not very sophisticated chassis/drivetrain/tire combination.