Question about fiding speed and distance

[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 got v=6 and distance=36m are these values correct?

 

[HallsofIvy]

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?

 

[sofiasherwood]

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]

Halls, it doesn't say constant acceleration. Constant power is what you might get from a car with a continuously variable transmission.

 

[sofiasherwood]

I am assuming the acceleration is constant. Is that right?

 

[sofiasherwood]

Have I gone wrong somewhere?

 

[Enigman]

Me thinks...
P=Fv
12=20*x'*x''
3/5x'=x"
x(0)=0, x'(0)=0
http://www.wolframalpha.com/input/?i=3/5=x"*x',+x(0)=0,+x'(0)=0
x=2(2/5)^1/2t^3/2
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]

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]

If we disregard x(9) value in question { http://www.wolframalpha.com/input/?i=3/5=x"*x',+x(0)=0,+x'(0)=0,x(9) }
For x(6) we get 24/(5)^0.5
http://www.wolframalpha.com/input/?i=3/5=x"*x',+x(0)=0,+x'(0)=0,x(6)

 

[adjacent]

Try not to round of numbers.Use fractions.

 

[Enigman]

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]

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.