# Simple Kinetics Q - 1 formula vs another

Discussion in 'MCAT Study Question Q&A' started by amusedtodeath24, May 28, 2008.

1. ### amusedtodeath24

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If you have a question that reads -

"A particle starts from rest and accelerates for 4 seconds at 10 m/s^2. What is its displacement from its initial position?"

Why is it that you cannot solve the question this way:

delta V = displacement / time
==> delta V * time = displacement?

(Since we are not given Vf, I solved for it by using Vf - Vi = a * delta t)

Instead, why is it that I have to use delta x = 1/2 a * t^2?

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3. ### nikeshp Banned

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both would work, i think you just got your equation wrong.
delta v = displacement / time is wrong i think.

it should be:

avg v = displacement / time

avg v = (Vf+Vi)/2

You should get 80 with both, given your numbers

4. ### amusedtodeath24

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Thank you so very much, I did have the equation wrong. Can I ask another one? (I am severely physics-challenged...)

"A particle travels 3.5 m. If its initial velocity is 9 m/s and its final velocity is 16 m/s, what is its acceleration?"

Using delta V^2 = 2 * a * delta x, I get 7, which is apparently not the correct answer.

The answer given is 25, which is obtained by using the aforementioned average velocity equation to get delta t, which is then plugged into delta V = a * delta t.

Am I missing something again? I am so confused. Thanks again.

EDIT -- I just realized my stupid mistake. It's not delta V ^2 ... I'm trying to be a smart ass and trying to combine Vf^2 - Vi^2 into delta V^2, which is not the same thing...well it is, but instead of squaring both numbers then subtracting, I'm subtracting THEN squaring. WRONG! AHH sorry

5. ### unsung

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Really, I would just use xf = xi + (vi)t + (1/2)at^2. If the problem gives you all the variables you need, why solve for another variable they don't ask for, and then use that to get the one they ask for? It saves time to use the best formula for the problem.

For all the kinetics stuff, I use one of the 3:

xf = xi + (vi)t + (1/2)at^2

vf = vi + at

(vf)^2 = (vi)^2 + 2ax

For the longest time, I only used the top 2, and would always derive the 3rd one if a problem called for it. Then I took the time to memorize the 3rd one and it shaved a lot of time off of how long it took me to get the answer.

6. ### amusedtodeath24

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Maybe I just am having a hard time seeing which equation to use when - for example:

"A particle moving at 5 m/s reverses its direction in 1 s to move at 5 m/s in the opposite direction. If its acceleration is constant, what distance does it travel?"

Here I am thrown off completely because I know if I use one of the kinetics eqs, I get displacement, which is not what they are asking for in this situation. How would I find distance here?

7. ### nikeshp Banned

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i know you would have to break it up into two problems, one for the distance it travels forward and the other for the distance it travels in the opposite direction.

how to do that, with the information given im not really sure. how long was the particle moving?

8. ### amusedtodeath24

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That's the problem verbatim; there's actually no further information given..and the solution makes absolutely 0 sense:

"x = vit + 1/2at^2 where a = deltav/t. But there is a trick. x is displacement, not distance. To find distance, you must recognize that the particle travels half the distance in half the time. Thus, you plug 1/2 the total time, only in the first equation and then solve for x and double your answer for the total distance."

Honestly...what the F*** did that just say? The answer is apparently 2.5 m. ?????

9. ### unsung

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Well, if they're asking for "distance" instead of "displacement", I would split the problem up into two parts: when the particle is moving in one direction (the initial direction, let's call it the + direction), and when the problem is moving in the opposite direction (the - direction).

So, by the wording of the problem, it's a bit ambiguous as to whether "1 second" refers to the time it takes for the particle to go from 5 m/s to -5 m/s OR the time it takes for the particle to simply "reverse direction", i.e. go from 5 m/s to 0 m/s.

Assuming it's the latter, it's easy. We'd just set it up:
vi = 5 m/s, vf = 0 m/s. (This is because the acceleration is in the opposite direction of its initial velocity, so at some point, the particle is going to stop and start to reverse directions.) So, we get:

vf = vi + at

0 = 5 + a(1)

a = -5 m/(s^2)

Plugging into:

vf^2 = vi^2 + 2ax

0^2 = 5^2 + 2(-5)x

x = (-25)/(-10) = 5/2 = 2.5 m

So now, we need to find the distance it travels starting from here at vi = 0, to vf = -5 m/s. An important thing to understand here is that it takes the same amount of time to move from vi = 5 m/s to vf = 0 m/s, as it does from vi = 0 m/s to vf = -5 m/s, when the particle is under the same constant acceleration.

An example of this is, let's say you're on a cliff, and you throw a ball up with a certain vi, let's say vi = 5 m/s (up is +). When the ball comes back down to the same height (i.e. height of the cliff), its velocity will have the same magnitude as the initial velocity, but the opposite direction, i.e. vf = -5 m/s (pointing down). (Here the constant acceleration is of course due to gravity.) You can prove this to yourself using the kinematics equations. But after that, just remember that fact... it comes in handy.

Therefore, the total distance the particle has traveled is 2 x 2.5 m = 5.0 m. The total displacement, on the other hand, is 0 m. Think of the cliff problem. If it's not so clear, I'll try to explain the cliff thing better.

10. ### nikeshp Banned

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that certainly seems like it makes sense, but the answer is apparently 2.5, not 5.0. i have no explanation for that, sorry.

11. ### nikeshp Banned

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unless the original assumption was wrong. if you find acceleration when vi = 5, vf = -5, and t = 1, you get a = -10 (sorry for not including units, but you know what they are) then if you solve for that doing the same thing unsung did, that would give you 2.5 i think.

12. ### amusedtodeath24

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thank you both very much for your help. unsung, i understood your explanation, and as nikeshp said it does make sense - however, the answer to the question is listed as 2.5 rather than 5.0. i understand how you got 5.0 for the distance and 0 as the displacement, but i definately do not understand how they got 2.5 as the distance. i wonder if it is an error in the book?

13. ### fileserver

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If you find the first half of the distance, a should be opposite sign of v, because the accerelation is changing v direction. If you plug v:5 then a:-10 . T is .5 bc 1s to change from -5 to 5.
The second half make more sense. Vi: 0 and a: 10. Plug into the equation and you should get 1.25 each part.

14. ### amusedtodeath24

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thank you very much guys.

splitting the problem into 2 parts -

the first part with vi = 5 and vf = 0, and t = 1/2
the second part with vi = 0 and vf = -5, and t = 1/2

by using delta x = vit + 1/2 at^2, i get 1.25 just like fileserver said

..however..

why is it that you get a completely wrong answer by using vf + vi/2 = delta x / delta t?
i guess i don't understand why the delta x = vit...etc. works and the average velocity equation doesn't...

ah well..thank you