Position and Speed of a Hockey Puck with Applied Force

[neurotikfisch]

Homework Statement

A hockey puck with mass 0.160 kg is at rest at the origin (x = 0) on the horizontal, frictionless surface of the rink. At time t = 0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; he continues to apply this force until t = 2.00 s. (a) What are the position and speed of the puck at t = 2.00 s? (b) If the same force is again applied at t = 5.00 s, what are the position and speed of the puck at t = 7.00 s?

Homework Equations

F=ma
v=d/t
d=at^2

The Attempt at a Solution

I figured first I would need to find the acceleration of the puck since both F and the mass are given, and I would need acceleration to figure out the distance. After plugging .250 N and 0.160 kg into the force equation, I came up with:

a = .250 N / 0.160 kg
a = 1.56 m/s^2

After finding the acceleration, I plugged that and t = 2.00 s into the equation for distance:

d = 1.56 m/s^2 * 2.00 s
d = 3.12 m

After finding the distance, I plugged it (and time) into the equation; v = d/t to find my speed:

v = 3.12 m / 2.00 s
v = 1.56 m/s

I know this can't be right, and now I'm completely lost. I wouldn't even know where to begin with part (b)! Would anybody be kind enough to point me in the right direction? :] Thanks!

 

[neurotikfisch]

Well I think I'm a little closer.

v_f = v_i + at is what I feel like I'm going to need, but I'm just not sure how to integrate acceleration here.

 

[mateomy]

You might want to try using a different equation. (I don't want to give TOO much away.) Good luck.

 

[neurotikfisch]

You might want to try using a different equation. (I don't want to give TOO much away.) Good luck.

I don't understand why. I've got all my variables accounted for in the equations I have, don't I? Am I not just going about the wrong way interpreting the problem?

 

[mateomy]

Is your answer not matching up with the book's? If you find the acceleration, you can plug it into find the position, then again plug it into find the velocity...not necessarily all the same equations. If your answer isn't matching up...to what extent is it off by? Sometimes that can give you a clue...Hope that provides a modicum of assitance.

 

[mateomy]

...also, check your velocity vs. acceleration; they shouldn't be the same number.

 

[neurotikfisch]

d = .5(v_o + v)t

I think I found it, but is this equation only applicable because the acceleration is constant? Maybe I'm not clear on why the first equation to find the distance wasn't relevant.