Kinematic Equation Confusion!

dolpho

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.

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?

dolpho

Couldn't I just move B over and then divide by C

a-b / c = x

Doc Al

Exactly. But I'd write it as x = (a-b)/c.

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

PhanthomJay

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.

dolpho

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?

Doc Al

Right.

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