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Simple velocity/acceleration/distance equation-but is mastering physics wrong?

by marewrath
Tags: equationbut, mastering, physics, simple
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marewrath
#1
Dec10-09, 05:46 PM
P: 5
A rock climber stands on top of a 57 -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 apart and observes that they cause a single splash. The initial speed of the first stone was 1.6 .

i will give you time, they both hit the water 3.3 seconds after the first stone is released (this both mastering physics and I agreed upon)

1. the initial speed of the second stone
2. the speed of the first stone as it hits the water
3. the speed of the second stone as it hits the water

so actually completed the questions, but the answers I got were different from the mastering phsyics answers--and I can't figure out how they got them. so if someone can scan these problems real quick and shoot their answers, id like to see if its mastering phsyics or me that is wrong.
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rl.bhat
#2
Dec10-09, 06:54 PM
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P: 4,439
Hi marewrath, welcome to PF.
Show your calculations, so that we can check it.
marewrath
#3
Dec11-09, 11:22 AM
P: 5
okay - i guess I jut wanted to check other peoples answers against mine versus masteringphysics, but here goes:

1. initial speed of the second stone:

57=vi+.5(9.8)(2.3)^2
57=vi+25.921
vi=31.079

2. speed of second stone as it hits the water

vf = 31.079 + 2.3(9.8)
vf=53.619
(of course, this is dependent on the answer from the last question)


3. the speed of the first stone as it hits the water
vf=1.6+3.3(9.8)
vf=33.94

cepheid
#4
Dec11-09, 11:38 AM
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Simple velocity/acceleration/distance equation-but is mastering physics wrong?

Quote Quote by marewrath View Post
A rock climber stands on top of a 57 [furlong? light year?] -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 [hours? orbital periods of Venus?] apart and observes that they cause a single splash. The initial speed of the first stone was 1.6 [c? knots? ] .

The point I'm trying to make by inserting some ridiculous, unrealistic units in the places where you left them out is that units are very important and you should always always include them in both the problem statement and in every step of your calculation. We don't tell you that just for the sake of being pedantic. Having the units in there helps you make certain your equations are dimensionally consistent (i.e. type of quantity on lefthand side = type of quantity on righthand side). If your units don't work out in this way, then you know you've made a mistake, so units are a valuable diagnostic. If you had done that, you would have caught the error you made below

Quote Quote by marewrath View Post
okay - i guess I jut wanted to check other peoples answers against mine versus masteringphysics, but here goes:

1. initial speed of the second stone:

57=vi+.5(9.8)(2.3)^2

Well, here is the problem. The equation is supposed to be:

d = vit + (1/2)at2

Without that factor of t in there, your equation is not dimensionally consistent, because you have:

[length] = [velocity]

which is nonsense. That's why you should always include the units in the calculation steps and make sure they work out.
marewrath
#5
Dec11-09, 11:45 AM
P: 5
i tried that originally and still wasnt getting the correct answers, but the ones without t times vi were closer so i figured that was probably more right and that t pertained to how long it was just going for that velocity... shrug
cepheid
#6
Dec11-09, 11:55 AM
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Quote Quote by marewrath View Post
i tried that originally and still wasnt getting the correct answers, but the ones without t times vi were closer so i figured that was probably more right
There is only one equation that is correct here, and it is the one I quoted for you. I would have hoped that you would have been convinced by the fact that your equation is not dimensionally consistent, rendering it meaningless! I would also hope that it would be clear to you that you can't just arbitrarily change the form of an equation and expect it to still be correct.


Quote Quote by marewrath View Post
and that t pertained to how long it was just going for that velocity... shrug
Nope. First of all, it's only "going at that [initial]" velocity for a single instant (i.e. for a time interval of zero width), so that doesn't make much sense. Second, If 't' in the first term of the equation was supposed to have meant something different from 't' in the second term of the equation, then we would have used a different symbol in order to make that explicitly clear. Same symbol means same quantity. In this case, t is meant to be the total time spent falling. To make this clearer:

vit - is the distance the rock *would* travel in time t if it started from speed vi and if it were NOT accelerating.

(1/2)at2 - is the distance the rock would travel in time t if it started from *rest* and were accelerating at 'a.'

In the most general case, you add these two distances together to get the total distance travelled.
marewrath
#7
Dec11-09, 12:13 PM
P: 5
hey hey! I wasn't saying I didnt believe you, or that I thought you were wrong at all, you are obviously correct; I was essentially saying that I confused myself early on, and thus began doing things wronger as I worked
cepheid
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Dec11-09, 12:19 PM
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Quote Quote by marewrath View Post
hey hey! I wasn't saying I didnt believe you, or that I thought you were wrong at all, you are obviously correct; I was essentially saying that I confused myself early on, and thus began doing things wronger as I worked
Alright I understand. Sorry if I seemed...animated. I just wanted to ensure you understood where you went wrong. Could you post what you got using the correct equations? It's possible that you just had a calculation error in there somewhere. I've done it and check against mine...


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