# Question about fiding speed and distance

by sofiasherwood
Tags: distance, fiding, speed
 P: 17 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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P: 39,310
 Quote by sofiasherwood 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 $m/s^2$. 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 $m/s^2$, then after time, t, it will have speed $at$ m/s and will have moved distance $(1/)at^2$ 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?
 P: 17 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
 Mentor P: 22,231 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.
 P: 17 I am assuming the acceleration is constant. Is that right?
 P: 17 Have I gone wrong somewhere?
 Best Humor P: 555 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...
P: 17
 Quote by Enigman 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.
 Best Humor P: 555 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
 PF Gold P: 1,459 Try not to round of numbers.Use fractions.
Best Humor
P: 555
 Quote by sofiasherwood 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...
 Sci Advisor PF Gold P: 2,241 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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