A particle of mass m is initially at rest

[Calpalned]

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.

Homework Equations

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?
http://photo1.ask.fm/726/204/412/-69996997-1sha200-dqjla0edmhpamrf/original/IMG_4066.jpg

 

[haruspex]

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

 

[Calpalned]

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

 

[haruspex]

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

Yes.

 

[Calpalned]

Thank you so much

 

[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?

 

[haruspex]

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 $\int t^2.dt$?

 

[Colin R]

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

I figured it out, it is the antiderivative I believe

 

[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.

 

[haruspex]

tell me if it's correct

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.