# Easy question newtons 1st law of motion, applying force to a hockey puck

1. Oct 3, 2009

### offbeatjumi

** sorry. this would be newtons 2nd law of motion. oops.

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

Hockey puck with mass 0.160kg at rest at x=0 and t=0 on horizontal frictionless surface. hockey player applies force of 0.250 N until t = 2.00s. ... ... if the same force is applied again at t = 5.00 s what is the position and speed of the puck at t = 7.00s?

2. Relevant equations

x(final) = x(initial)+v(initial)t+(1/2)a(t^2)
v(final) = v(initial)+at

3. The attempt at a solution

at t = 2 seconds, the puck is at x = 3.125m and the velocity is 3.125m/s. the acceleration (from F = ma) is 1.5625 m/s^2.

how do I incorportate reapplying the force? i imagine since the surface is frictionless and the acceleration is constant that the puck keeps increasing in velocity right?

Last edited: Oct 3, 2009
2. Oct 4, 2009

### tiny-tim

Hi offbeatjumi!

(try using the X2 tag just above the Reply box )
I'm not sure what you're asking.

The puck will have zero acceleration, and therefore the same velocity, until the force is reapplied.

So find the new x at the end of that constant velocity, and use that as the initial x in the same equation as before.

3. Oct 4, 2009

### offbeatjumi

im sorry i worded that wierdly

this is essentially the question: if the same force is applied again at t = 5.00 s what is the position and speed of the puck at t = 7.00s?

but what you said helps, ill try it