# A particle of mass m is initially at rest

## Homework Statement

A particle of mass m, initially at rest at x=0, is accelerated by a force that increases in time as F=Ct2. Determine its velocity v as a function of time.

x = vt
v = at

## The Attempt at a Solution

The correct method makes sense, but my method has no error. However, my answer is wrong. Why?

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haruspex
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v=at only works for constant acceleration. Integration works for constant or varying acceleration.

So because force is a function of time (f = Ct2), acceleration is also a function of time and therefore not constant?
Thanks

haruspex
Homework Helper
Gold Member
2020 Award
So because force is a function of time (f = Ct2), acceleration is also a function of time and therefore not constant?
Thanks
Yes.

Thank you so much

I don't see how you got 3m at the bottom? does it have to do with the fact that t is to the third power?

haruspex
Homework Helper
Gold Member
2020 Award
I don't see how you got 3m at the bottom? does it have to do with the fact that t is to the third power?
Yes. What is ##\int t^2.dt##?

Yes. What is ##\int t^2.dt##?
I figured it out, it is the antiderivative I believe

I solved this questions like;

$$F=m.a$$ $$F=Ct^2$$ => $$m.a = Ct^2$$ $$a= \frac {Ct^2} {m}$$
and if we integrate "a" respect to time(t), we obtain velocity(v);
$$\int \frac {Ct^2} {m} dt = \frac {Ct^3} {3m} = V$$
and if we integrate "v" respect to time(t), we obtain position(x);
$$\int \frac {Ct^3} {3m} dt = \frac {Ct^4} {12m} = X$$

I'm not sure, tell me if it's correct.

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haruspex