Finding Distance -- Hockey Puck Velocity

[Catchingupquickly]

Homework Statement

An ice hockey puck leaves a hockey stick with a velocity of 45 m/s, how far will it travel in 3.0 seconds?

Homework Equations

D = v/t or

$\Delta \vec d= \frac 1 2 \vec a \Delta t^2$

with $\vec a = v_2 / \Delta t$

The Attempt at a Solution

[/B]
D = v/t
= 45 m/s / 3 = 135 meters

or $0.5 (15 m/s^2) (3)^2$ = 67.5 meters

Which one is it, and more importantly... how do I tell the difference on when to use each formula?

Thank you

 

[kuruman]

D = v t.

https://www.physicsforums.com/threads/introductory-physics-formulary.110015/

 

[TomHart]

Since no friction is mentioned, I believe for this problem you should assume there is no friction, which means the velocity is constant. In other words, a = 0.

 

[Catchingupquickly]

Thank you both very much. Very helpful.

 

[Catchingupquickly]

Actually a follow up question to this adds friction. The puck is hit with a force of 15.3 N and the friction slowing it down is 0.75 N.
Same time (3.0 s) same velocity (45 m/s)

So a = f/m then plug that into the second formula mentioned above?

 

[haruspex]

$\Delta \vec d= \frac 1 2 \vec a \Delta t^2$
with $a = v_2 / \Delta t$

Those equations are not quite right. It is $\Delta d= v\Delta t+ \frac 1 2 a \Delta t^2$ and $a =\Delta v/\Delta t$.
You are not told the puck stops in 3 seconds. If you assume it keeps going at v then Δv=0 and a =0, so you end up with the same equation as d=vt.

 

[Catchingupquickly]

Extremely helpful. Thank you