# Kinematics - Constant Acceleration

1. Sep 23, 2009

### JG89

1. The problem statement, all variables and given/known data

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.

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?

2. Relevant equations

3. The attempt at a solution

I have three motion equations for constant acceleration. But I have no idea where to start this problem...I know it's not going to be a numerical answer. The answer will be expressed in terms of my variables.

Last edited: Sep 23, 2009
2. Sep 23, 2009

### TVP45

Is there a question in there?

3. Sep 23, 2009

### JG89

Oops. I just edited my original post with the question included.

4. Sep 23, 2009

### TVP45

Are there any actual numbers in there?

5. Sep 23, 2009

### JG89

Nope

6. Sep 23, 2009

### TVP45

Not even for "last instant"? What do you suppose that means?

7. Sep 23, 2009

### JG89

Not even for "last instant"...

I suppose that means that my answer will be expressed in variables...If the drag-racer waits until the last instant to hit the pedal, then shouldn't my car and the race-car's position be virtually the same at t = 0?

If at t-max our positions are again the same, then it seems like I am going to have to use the equation $$s_f = s_i + v_i(t_{max} - t_0) + 0.5a(t_{max} - t_0)^2$$, as s_f will have two solutions (one at t = 0 and one at t-max) and I will probably be solving a quadratic...

8. Sep 23, 2009

### JG89

Okay, I think I've got it.

For my car: $$s_f = s_i + v_i(t_{max} - t_0)$$. The question said to put t_0 = 0, and I also know that s_i = 0, so the equation simplifies to $$s_f = v_i(t_{max})$$, where v_i is the constant velocity that I'm traveling at (it's not 0).

Now, I want to know when the drag car is also at position s_f. So the equation for the drag-car is: $$s_f = s_i + v_{i-drag}(t_{max} - t_0) + 0.5a(t_{max} - t_i)^2$$.
v_i-drag is the drag-cars initial velocity, which is 0. Also remember that t_i = 0. s_i also equals 0. So the equation simplifies to $$s_f = 0.5a(t_{max})^2$$.

Setting the two equations equal to eachother: $$0.5a(t_{max})^2 = v_i(t_{max}) \Leftrightarrow -0.5a(t_{max})^2 + v_i(t_{max}) = 0$$. Using the quadratic formula, I have $$t_{max} = \frac{-v_i \pm v_i}{-a}$$. There are two solutions to this. When $$t_{max} = 0$$ and when $$t_{max} = \frac{2v_i}{a}$$. I will take the second solution.

Is this correct?

9. Sep 23, 2009

### TVP45

I suppose you might state "last instant' as referring to some distance before the eminent collision (and that could as easily be some time), so you might decide that the "chase car" is travelling at Vo and the dragster accelerates when the chase car is -x behind it. The quadratic implies you would be coincident twice and that can't happen if you're on a single track. So, I think you have to consider the first collision and you might as well let that happen down at the traps.

10. Sep 23, 2009

### JG89

So then can't I take my original answer of t_max = 2v_i/a and just subtract a tiny bit of time from it? So t_max = 2v_i/a - t, where t is a small bit of time?

11. Sep 24, 2009