# Kinematics / ODEs problem

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1. Aug 15, 2015

### Remixex

1. The problem statement, all variables and given/known data
I stumbled upon a problem and i can't establish the ODE to solve it, from there on i believe i can solve the ODEs if they have regular analytical solving methods (translated from Spanish, will sound a bit weird)
Car race, 2 pilots (a and b) participate in a drag race. They begin their movement at rest and then accelerate at a constant rate.
Car b
finishes the last quarter of the way in 3 seconds, car a finishes the last third of the way in 4 seconds
Who wins and by how long?
Secondary question, what does "constant rate of acceleration" mean?, Line with or without slope? i believe it's with slope, or else we wouldn't need a differential equation, constant rate should mean constant derivative.

2. Relevant equations
Differential form of basic Kinematics equations

3. The attempt at a solution
So i get X(t) represents function of distance, X'(t) a function of velocity and X''(t) one of acceleration, also we know that X(0)=0 (therefore zero velocity and acceleration at t=0) for both cars. I believe we need to establish 2 differential equations for each car? i'm not sure
Also, if they accelerate at a constant rate, X''(t) may have the form of mt+c

2. Aug 15, 2015

### Qwertywerty

Constant rate of acceleration implies X''(t) is a constant .

3. Aug 15, 2015

### brainpushups

This is strikingly similar to a bonus question I assign to my introductory physics class except that it doesn't involve the cars traveling at their max speed after they accelerate. Honestly I think that resorting to differential equations is a bit overkill. If you know the acceleration is constant just invoke the kinematic equations (derive them from x''(t) = a if you wish - not a bad exercise if you're just learning about them).

4. Aug 15, 2015

### Remixex

I thought of the same, normal Kinematics equations hold water if the acceleration is a constant, i had this idea in my mind the acceleration was a linear function, as in mt+c, but well if it isn't then normal equations are usable, and no i just stumbled upon this question on a forum and it caught me off guard, i know about ODEs :D

5. Aug 15, 2015

### HallsofIvy

Staff Emeritus
It's hard to tell if you really do not understand what you are doing or if you are just saying it badly!

Acceleration of an object is a physics property and does not have anything to do with a graph so your "Line with or without slope?" makes no sense.

"constant acceleration" means that the velocity is changing at a constant rate. I'm surprised you did not know that.

I'm not sure whether you really do not understand or are just wording this incorrectly. X'(t) and X''(t) are both "functions of" t- that's what the "(t)" means! What you mean to say, I think, is that X'(t) is the velocity function and X''(t) is the acceleration.

No. The fact that X(0)= 0 means that both cars position at t= 0 is 0. That says nothing about their velocity or acceleration at t= 0.
"They begin their movement at rest"
tells you that the velocity is 0 at t= 0. The acceleration at t= 0 is certainly NOT 0. Your are told the acceleration is a constant and if that constant were 0 neither car would ever move!

No, if they accelerate at a constant rate, X''(t) has the form "X''(t)= a" where a is that constant. Integratig, the velocity, X'(t)= at+ c.
To get X(t) for each car integrate that again.
(But you are NOT told that the constant integration is the same for both car so use different letters for the acceleration.

6. Aug 15, 2015

### Ray Vickson

"Constant acceleration" would definitely mean that $x''(t) = \text{const.}$, but putting in the word "rate" makes it less clear. I think one could, in all honesty, interpret it to mean that $x'''(t) = \text{const.}$, by analogy with other cases where 'rate' stands for a time-derivative. If this interpretation is correct there would still be no need to solve a DE, because integrating twice would suffice. However, in that case there might not be enough information given to allow a solution of the problem.