# Homework Help: A particle of mass m is initially at rest

1. Feb 8, 2015

### Calpalned

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
x = vt
v = at

3. The attempt at a solution
The correct method makes sense, but my method has no error. However, my answer is wrong. Why?

Last edited by a moderator: May 7, 2017
2. Feb 8, 2015

### haruspex

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

3. Feb 11, 2015

### Calpalned

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

4. Feb 11, 2015

### haruspex

Yes.

5. Feb 11, 2015

### Calpalned

Thank you so much

6. Mar 3, 2016

### Colin R

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?

7. Mar 3, 2016

### haruspex

Yes. What is $\int t^2.dt$?

8. Mar 3, 2016

### Colin R

I figured it out, it is the antiderivative I believe

9. Mar 4, 2017

### mastermechanic

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.

Last edited: Mar 4, 2017
10. Mar 4, 2017

### haruspex

It is, but technically you should go through the step of allowing for a constant of inrtegration each time, then using the initial conditions to show it is zero.