- #1

- 34

- 0

## Homework Statement

*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.

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.

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?

## Homework Equations

[PLAIN]https://www.physicsforums.com/latex_images/90/905663-4.png [Broken]

## The Attempt at a Solution

So T-max is the amount of time from when the dragster accelerates at the last moment and both the accelerating dragster and constant velocity car hit or nearly touch each other correct? That's my main issue with the question. If so...

The car would be given an equation of x = -x_0 + v_0t as it does not accelerate and has a position negative to the dragster which would be the origin correct?

Then the dragster would have an equation for x = .5at^2 as it is at the origin and is accelerating.

To reach t-max these would need to be touching each other so I can equal one to another and then weed out t correct?

Doing so I get 0 = .5at^2 - v_0t + x_0 and would then would use the quadratic equation to simplify which becomes t = (v_0 +/-(sqrt(v_0^2-2ax)))/a.

Now I'm looking at the answer and it says t-max = v_0/a and that's not quite what I've got. Am I on the right trail or completely off?

EDIT: Just a though, would 2ax become 2v_0^2 as m/s^2 * m would give me m^2/s^2 which is velocity squared? Even with that I still don't get the answer but just a thought that I had.

Last edited by a moderator: