Question about fiding speed and distance


by sofiasherwood
Tags: distance, fiding, speed
sofiasherwood
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#1
Nov11-13, 07:58 AM
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I had a test earlier today and one question was,

A 20kg object initially at rest is accelerated at constant power of 12.0w. After 9.0s it has moved 56.0m. Find its speed at t=6.0s and its position at that instant.

I got v=6 and distance=36m are these values correct?
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HallsofIvy
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#2
Nov11-13, 08:26 AM
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Quote Quote by sofiasherwood View Post
I had a test earlier today and one question was,

A 20kg object initially at rest is accelerated at constant power of 12.0w. After 9.0s it has moved 56.0m. Find its speed at t=6.0s and its position at that instant.
I started to say that this makes no sense. An object accelerates at a constant acceleration measured in [itex]m/s^2[/itex]. That acceleration may be caused by a constant force, measured in Newtons, but not a constant power measured in watts.

But the "12.0 w" really has nothing to do with the problem and can be ignored. If an object accelerates from rest at constant acceleration, a [itex]m/s^2[/itex], then after time, t, it will have speed [itex]at[/itex] m/s and will have moved distance [itex](1/)at^2[/itex] m.

I got v=6 and distance=36m are these values correct?
They do not look at all right to me. How did you get them?
sofiasherwood
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#3
Nov11-13, 08:56 AM
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P*t=W
12*9=108
therefore W=108joules
W=F*displacement
F=108/56
F=1.93N

F=ma
a=1.93/20
a=0.1

so at t=6
Pt=W
12*6=72joules
W=72joules

W=F*displacement
displacement=72/(20*0.1)
displacement=36m

P=Fv
v=12/(20*0.1)
v=6

russ_watters
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#4
Nov11-13, 08:57 AM
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Question about fiding speed and distance


Halls, it doesn't say constant acceleration. Constant power is what you might get from a car with a continuously variable transmission.
sofiasherwood
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#5
Nov11-13, 08:58 AM
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I am assuming the acceleration is constant. Is that right?
sofiasherwood
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#6
Nov11-13, 09:01 AM
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Have I gone wrong somewhere?
Enigman
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#7
Nov11-13, 09:07 AM
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Me thinks...
P=Fv
12=20*x'*x''
3/5x'=x"
x(0)=0, x'(0)=0
http://www.wolframalpha.com/input/?i...%27%280%29%3D0
x=2(2/5)1/2t3/2
pugging t=9
x=18(6/5)^1/2
whereas according to question it should be 56m???
Either the question, me or wolframalpha is incorrect...
sofiasherwood
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#8
Nov11-13, 09:12 AM
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Quote Quote by Enigman View Post
Me thinks...
Either the question, me or wolframalpha is incorrect...
This is the question given in the test word for word. The question confused me for ages, in the end I just assumed acceleration was constant. I don't know how you could work it out any other way.
Enigman
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#9
Nov11-13, 09:41 AM
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If we disregard x(9) value in question { http://www.wolframalpha.com/input/?i...3D0%2Cx%289%29 }
For x(6) we get 24/(5)^0.5
http://www.wolframalpha.com/input/?i...3D0%2Cx%286%29
adjacent
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#10
Nov11-13, 10:17 AM
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Try not to round of numbers.Use fractions.
Enigman
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#11
Nov11-13, 10:24 AM
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Quote Quote by sofiasherwood View Post
I am assuming the acceleration is constant. Is that right?
Acceleration can not be constant...
P=dW/dt
P=d(∫F.dx)/dt
P=F.dx/dt
P=Fv
If a is const.
P=ma*(at)
P=ma2t
Then Power is not constant as given in the question...
f95toli
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#12
Nov12-13, 04:52 AM
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I saw this question when it was in the other section, it looked fishy to me then and I just had a go at it. Unless I am missing something it does not have a solution.
The way the question is formulated you will -as Enigman pointed out- end up with a 2nd order ODE for the position; but then the constants are over-determined.

I wonder if this is a case of the teacher trying to add a red herring (constant power) without realizing that this implies that the accelaration can not be constant.


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