How Long Can You Avoid a Collision with an Accelerating Dragster?

[aleferesco]

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 v₀ 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?


please help me, just get me started . I've look this question from all possible angles but couldn't figure out how to start

Homework Equations

v_f = v_i +a(delta)t

s_f = s_i +v_i(delta)t + 1/2a(delta)t^2

v_f^2= v_i^2 +2a(delta)s

ps. "s" is position
"V" velocity
"f" final
"i" initial
I use the word delta instead of the little triangle, I don't know how to put it

The Attempt at a Solution

My first attempt was trying to figure out the constant velocity of the car, but the thing is that there's not enough given values in order to solve (or are there enough?)


Am I using the wrong formulas?

Any kind of help is truly appreciated

 

[Redbelly98]

The condition to use is that s_f is the same for the car and the dragster. So I'd work with that equation.

 

[aleferesco]

oo true man, thanks I'll give it a shot

 

[aleferesco]

so I gave the question a try and her is what I came up with

Equation for the accelerating dragster

(sf)_d = s₀ +v₀t + 1/2at²

Equation for the car going at a constant velocity

(sf)_c = s₀ + v₀t


then...

(sf)_d= (sf)_c

1/2at² = v₀t

at² = 2v₀t

at = 2v₀

t_max = 2v₀/a

 

[aleferesco]

please let me know if it is right or wrong or give me some pointers

 

[Redbelly98]

You're on the right track, but there is a detail or two we need to think about more carefully.

Equation for the accelerating dragster

(sf)_d = s₀ +v₀t + 1/2at²

Equation for the car going at a constant velocity

(sf)_c = s₀ + v₀t


then...

(sf)_d= (sf)_c

1/2at² = v₀t

Looks like you're saying v0 is zero for the dragster, and v0 is not zero for the car. That's correct. (It is potentially confusing since you used the same term, v0, for the car and for the dragster.)

It also looks like you're saying s0 is the same for the car and dragster. That would be a problem, since it means they have already collided at t=0.

 

[aleferesco]

so would this be better then

Equation for the accelerating dragster

(sf)_d = s₀)_d +(v₀)_dt + 1/2at²

Equation for the car going at a constant velocity

(sf)_c = (s₀)_c + (v₀)_ct


then...

(sf)_d= (sf)_c

1/2at² = (s₀)_c + (v₀)t

at² = 2(v₀t + (s₀)_c)


at² = 2v₀t + 2(s₀)_c

does it makes more sense now?

or I'm completely off

 

[Redbelly98]

Yes, that's correct. Proceed.

 

[aleferesco]

so if what I did was correct then now I have to Isolate for t,

at² = 2v₀t + 2(s₀)_c

t² = (2v₀t + 2(s₀)_c)/a

t² = (2(v₀t + (s₀)_c))/a


t = [(2(v₀t + (s₀)_c))/a]^1/2


but my question is, what can I do with the 't' that is inside of the equation

does it seem right?

 

[Redbelly98]

By the way, this is not a very "straightforward" problem ... as you might suspect by now.

Note we have a quadratic equation in "t":
a t² = 2 v0 t + 2 (s0)_c

And there is a formula for solving quadratics (think back to high school algebra) ...

 

[aleferesco]

ooo!

0 = at² - 2v₀t - 2(s₀)_c

t = (2v₀t +- [(-2v₀)² - 4(a)(-2(s₀)_c)]^1/2/2a

since time is only positive then... I only take the positive side of the obtained values..

t_max = (2v₀t + [4(v₀)² - 8(a)(s₀)_c)]^1/2/2a

 

[Redbelly98]

You're getting there.

ooo!

0 = at² - 2v₀t - 2(s₀)_c

t = (2v₀t +- [(-2v₀)² - 4(a)(-2(s₀)_c)]^1/2/2a

Get rid of the "t" on the right hand side, it's the variable we're solving for.

since time is only positive then... I only take the positive side of the obtained values..

t_max = (2v₀t + [4(v₀)² _- 8(a)(s₀)_c)]^1/2/2a

Uh, not so fast.
The minus-sign I've highlighted should be a +. And the "negative" part of the +- doesn't necessarily give a negative time, so we should keep it as +- for now.

 

[Redbelly98]

I've (hopefully) thought of a better way to explain what's going on.

A graph of the car's position is a straight line, with slope v0 and unknown y-intercept s0.

A graph of the dragster's position is a parabola, x = (1/2) a t^2

The two graphs can intercept at two values of t, given by the quadratic formula solution for t. A key thing to realize is that the earlier-time (lower t, "-√") solution is when the crash occurs. So take the "-√" solution.

After that, find the value of s0 which gives the largest value of t.