Unifom accelaration question about racing

AI Thread Summary
Go-cart A maintains a constant speed of 20.0 m/s, completing the 1.0 km track in 50 seconds. Go-cart B starts from rest and accelerates uniformly at 0.333 m/s². To determine the winner, the time taken by each go-cart to finish the race must be calculated using kinematic equations. Go-cart B's time can be found using the formula for distance under constant acceleration, while Go-cart A's time is straightforward due to its constant speed. The calculations reveal which go-cart wins and by how much.
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Homework Statement


Two go-carts, A and B, race each other around a 1.0km track. Go cart A travels at a constant speed at 20.0m/s . Go cart B accelerates uniformly from rest at a rate of 0.333m/s2. Which go-cart wins the race ? and by how much ?

EDIT: When it says around it means they raced one lap around the track going at a constant speed, except for Cart B, its going at a constant acceleration

Don't go too deep into the shape of the track, it doesn't really matter, however if you still do, its going in a straight path.

Homework Equations


I don't know which to use.

The Attempt at a Solution



Well I know that Go-Cart A take 50 seconds to cross the line.

20m/s / 1000m = 50 seconds
 
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Do you mean to say that the track is not circular? After all, you said that
race each other around
, by "around", did you mean perhaps that the segment is round?
If not, recall that kinematics equations:
<br /> \large<br /> x = \frac{at^2}{2}<br />
Where a is your constant acceleration; But do clarify your stance on the shape of field etc...
Daniel
 
Fixed
 
Okay, thank's for allaying the mix-up.
What's stopping you from using the equation above? Find the time it took each to cross said distance, which will indicate firstly, who won... Then, measure how deficient/far was the loser by that moment in time.
Daniel
 
Don't know what equation to use ...

so lost.. physics teacher was away for a week and the whole class is lost
 
Okay, okay, not to worry, there's no need to panic.
You have, from my post:
<br /> \large<br /> x = \frac{at^2}{2}<br />
So plug in your data here for each car.
In one, x is simply v*t(the one with the constant velocity).
in the other, the above mentioned formula.
Find T for each(a is the constant acceleration).
Compare the results; Whosever t is greater, lost the race(took him longer to cross the track).
Then, find where was the loser at the time T of the victor.
It's as simple as that,
But it would help greatly if you could point-out where you're having difficulties,
Daniel
 
Thanks a lot Daniel, you're a life saver

MAY GOD BLESS YOUR WARM KIND HEART ! <33333333333333333
 
So I reckon you solved it properly, well done! :)
Good job,
And keep up the good work,
Daniel
 
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