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Kinematics - Drag Racer

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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.


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.
 
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Answers and Replies

  • #2
collinsmark
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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?
Actually, both are right! :smile: (Yeh!) It's just that your answer isn't quite finished yet. :frown: (Ohhh.)

The crux of the problem is embodied in, "The dragster driver sees you coming but waits until the last instant to put down the hammer."

In other words, x0 is as small as physically possible. Your equations (even your final one) still contain an x0 in them. Find the value of t that minimizes x0. :wink: (Hint: you may find it easier to use your first equation rather than your second [from the two equations that I quoted above.])
 
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Okay, so x_0 is the distance between the car and the dragster when the dragster takes off correct? So the last instant would suggest that x_0 is as small as it can be without the car crashing into the dragster. I need to find a t value(t-max) such that x_0 is as small as possible. That's the concept of this correct? I'm not quite understanding how I would minimize the x_0 in 0 = .5at^2 - v_0t + x_0
 
  • #4
collinsmark
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Okay, so x_0 is the distance between the car and the dragster when the dragster takes off correct? So the last instant would suggest that x_0 is as small as it can be without the car crashing into the dragster. I need to find a t value(t-max) such that x_0 is as small as possible. That's the concept of this correct? I'm not quite understanding how I would minimize the x_0 in 0 = .5at^2 - v_0t + x_0
First, do a little algebra solving for x0, thus expressing it as a function of t.

To maximize or minimize a function, take the derivative with respect to whatever variable you wish to vary, and then set the result equal to zero. Solve for the variable (algebra). That's the value of the variable that maximizes or minimizes the function.*

*The answer can give local maxima or local minima. So if you get multiple values (such as in a quadratic), you might wish to do some sanity checking to figure out what each value is. In this particular problem though, you end up with a single minimum (no maximum) so the need for sanity checking doesn't apply here.

The idea behind the procedure is this: When a smooth function rises to the top of its [possibly local] maximum or bottom of its [possibly local] minimum, the slope of the line is zero at this point. So the above procedure is merely finding where the slope of the function equals zero, as the variable changes.
 
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  • #5
verty
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This is one of those questions where a little bit of mathematical intuition can save a lot of work.

[tex]t = \frac{v_0 \pm \sqrt{v_0^2 - 2ax_0}}{a}.[/tex]

The intuition is that the minimum (sorry, maximum) occurs when the expression under the square root = 0. If that value is negative, the car crashes into the dragster (mistake, it means the car does not reach the dragster :biggrin:), if it is 0, they just reach each other.

If [tex]v_0^2 = 2ax_0[/tex], then t = v_0/a, as required.

Without this knowledge, one would have to follow the method set out above.

As for why this works, it is because the formula for t needs to reflect the fact that some configurations are impossible, and in that formula this is only possible when the number under the square root is negative.

-- Admittedly, this is a little advanced for introductory physics (in that it relies on mathematical intuition), so see this as a mathematical aside. But knowing things like this can be useful for spotting errors (as long as one doesn't make more errors :uhh:).
 
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  • #6
collinsmark
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This is one of those questions where a little bit of mathematical intuition can save a lot of work.

[tex]t = \frac{v_0 \pm \sqrt{v_0^2 - 2ax_0}}{a}.[/tex]
You could take the derivative of that function, solve for x0. That would give you the particular value of x0 that maximizes t. But you're not interested in the particular value of x0 that maximizes t! You're interested in the particular value of t that minimizes x0. So if you were to proceed that way, you'd have to continue substituting the resulting x0min expression back into the equation and solving for t. But there is a much easier way.

(You might want to re-read my previous post)

Start with your original equation,
0 = ½at2 - v0t + x0.​
Solve for x0. Now take the derivative of that function (with respect to t). That gives you the specific value of t that minimizes x0. :wink:
 
  • #7
verty
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Collinsmark said:
You're interested in the particular value of t that minimizes x0.
I agree with everything you've said, Collinsmark, except in one small detail. Consider this expression:

[tex]v_0^2 - 2ax_0.[/tex]

This expression gets more negative when x_0 gets bigger. In that sense, x_0 has a maximum possible value. Too big and the car is just too far behind the dragster to ever reach it. x_0 also has a minimum value, but that is just 0, which means the car reaches the dragster at time 0.

I agree that it seems that x_0 must be minimized, and it caught me out too. Luckily I saw by the form of the equation that it was the other way around.

This was not an easy question to figure out, by any means.
 
  • #8
verty
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Okay, I think I totally misunderstood the question. To be honest, it was because it was worded in a strange way.

The question says that the dragster accelerates at the latest possible moment, but I think the driver did that only to avoid crashing. What we must take from that is that the velocity of the dragster = v_0 when they meet.

Let t_i be the time the cars meet.

v_dragster = v_car
... (hidden)

PS. I think the teacher wanted a longer solution that this, using the derivative.

--------------------------------------------

Sorry Collinsmark, I didn't mean to hijack your thread (you were replying first).

Exuro89, just ask if you are confused about anything or have questions.
 
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  • #9
collinsmark
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Okay, I think I totally misunderstood the question. To be honest, it was because it was worded in a strange way.

The question says that the dragster accelerates at the latest possible moment, but I think the driver did that only to avoid crashing. What we must take from that is that the velocity of the dragster = v_0 when they meet.

Let t_i be the time the cars meet.

v_dragster = v_car
... (hidden)

PS. I think the teacher wanted a longer solution that this, using the derivative.
Well, it seems that you've found yet another way to approach this problem with the velocities being equal. :smile: Yes, it turns out that that's correct. When the car comes up just behind the dragster, to the point where they almost touch, their velocities are in fact equal at that moment. But this might not be so obvious as the fact that their positions are equal at that moment.

So yes, the teacher probably wants one to solve the problem with the assumption that at time t, their positions are equal, and not necessarily their velocities (even though the velocities turn out to be equal too, but let's ignore that).
 

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