simple velocity/acceleration/distance equationbut is mastering physics wrong?by marewrath Tags: equationbut, mastering, physics, simple 

#1
Dec1009, 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 answersand 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. 



#2
Dec1009, 06:54 PM

HW Helper
P: 4,442

Hi marewrath, welcome to PF.
Show your calculations, so that we can check it. 



#3
Dec1109, 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 



#4
Dec1109, 11:38 AM

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simple velocity/acceleration/distance equationbut is mastering physics wrong?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 Well, here is the problem. The equation is supposed to be: d = v_{i}t + (1/2)at^{2} 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. 



#5
Dec1109, 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




#6
Dec1109, 11:55 AM

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v_{i}t  is the distance the rock *would* travel in time t if it started from speed v_{i} and if it were NOT accelerating. (1/2)at^{2}  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. 



#7
Dec1109, 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




#8
Dec1109, 12:19 PM

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