Who Wins the Drag Race Based on Constant Acceleration?

[Remixex]

Homework Statement

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.

Homework Equations

Differential form of basic Kinematics equations

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

 

[Qwertywerty]

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

 

[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).

 

[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

 

[HallsofIvy]

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

Homework Statement

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?

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.

i believe it's with slope, or else we wouldn't need a differential equation, constant rate should mean constant derivative.

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

2. Homework Equations

Differential form of basic Kinematics equations

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[,

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.

also we know that X(0)=0 (therefore zero velocity and acceleration at t=0) for both cars.

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!

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

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.

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

Differential form of basic Kinematics equations

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

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

"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.