Kinematic Equation Confusion!

1. The problem statement, all variables and given/known data

Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

2. Relevant equations

3. The attempt at a solution

Finding V(final)= Vf-Vi = at ---> Vf = (-9.81)(2.6) = 25.5

The second part is a little confusing. Why don't these equations come out with the same answer?

Vf^2 = Vi^2+2a(DeltaX)...Rearranged to... -DeltaX = 0 + 2(9.81) - 25^2 = 650meters something which doesn't really make sense.

The other I used was X= 1/2at^2 = 33meters

The second is the correct answer but why didn't the equation I do work? Since we know the initial is 0, the final is 25.5. We also know the time.

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 Quote by dolpho Vf^2 = Vi^2+2a(DeltaX)...Rearranged to... -DeltaX = 0 + 2(9.81) - 25^2 = 650meters something which doesn't really make sense.
Your rearrangement is incorrect. Give it another shot.

Try this first: If you had an equation a = b + cx, how would you solve for x?

 Quote by Doc Al Your rearrangement is incorrect. Give it another shot. Try this first: If you had an equation a = b + cx, how would you solve for x?
Couldn't I just move B over and then divide by C

a-b / c = x

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Kinematic Equation Confusion!

 Quote by dolpho Couldn't I just move B over and then divide by C a-b / c = x
Exactly. But I'd write it as x = (a-b)/c.

Now do the same thing with your equation. They are similar.

 Recognitions: Gold Member Homework Help Science Advisor Your algebra is wrong . 25^2 = 0 + 19.62x, solve for x. You shouldn't be using that equation because if you got Vf wrong in part 1, then this part would also be wrong. You can use it as a check after you use the x = .5gt^2 equation.

 Quote by Doc Al Exactly. But I'd write it as x = (a-b)/c. Now do the same thing with your equation. They are similar.
Ok so, V^2 = U^2 + 2aD

D = -V^2 / -2A

Ohhhhh, heheheh oops lol. I totally knew that but I tried to rush it. Sorry I have one more question.

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

So the reason I can't use V= D / T is because that only calculates average or constant velocity? So instead I'd have to use x = xi + vit + .5at^2?

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 Quote by dolpho So the reason I can't use V= D / T is because that only calculates average or constant velocity? So instead I'd have to use x = xi + vit + .5at^2?
Right.

If you know how to relate average velocity (given by D/T) to the final velocity, you can use that method as well.