Find Acceleration with only Initial Velocity and Final position?

AI Thread Summary
The discussion revolves around calculating the acceleration of a hockey puck that comes to rest after sliding 238 meters from an initial velocity of 3.8 m/s. The key equation used is v² = (vo)² + 2a(x - xo), where the final velocity is zero. Participants clarify that the acceleration should be negative due to deceleration, leading to a calculated value of -0.030 m/s². There is confusion regarding the time calculation, with one participant initially arriving at an implausibly long duration of 125.3 seconds. The correct interpretation emphasizes that the puck is decelerating, confirming the negative acceleration value.
fireykitty
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Homework Statement



A hockey puck sliding on a frozen lake comes to rest after 238 m. If its initial velocity is 3.8 m/s what is its acceleration if it is assumed constant? Answer in units of m/s^2

How long is it in motion? What is its speed after traveling 180 m?

Homework Equations



x=xo + vot +1/2at^2

The Attempt at a Solution



v^2=(vo)^2+2a(x-xo)
...v^2=(3.8)^2 + 2a (-238)

but i don't know the velocity, cause i don't have the time!

3.8 m/s * 1/238 m = time?
I find that I can't do anything without the time! I'm so lost and confused. PLEASE HELP.
 
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Here's a hint: what does "comes to rest" mean? ;-)
 
fireykitty said:

The Attempt at a Solution



v^2=(vo)^2+2a(x-xo)
...v^2=(3.8)^2 + 2a (-238)

but i don't know the velocity, ...

Welcome to PF.

That equation works. Your final velocity is 0 so ...

v2 = 3.82 = 2*a*238

Then you have your acceleration a.

With that v = a*t so t becomes easy to find.

To find the speed at 180 m just use the equation you gave above to determine the speed at that distance.
 
LowlyPion said:
Welcome to PF.

That equation works. Your final velocity is 0 so ...

v2 = 3.82 = 2*a*238

Then you have your acceleration a.

With that v = a*t so t becomes easy to find.

To find the speed at 180 m just use the equation you gave above to determine the speed at that distance.

But then we have two variables, a, and v, how do I solve then? it doesn't work if you manipulate and substitute v =at because then the other variable, t, comes in.
 
fireykitty said:
But then we have two variables, a, and v,
No you don't, you only have one, a. Look carefully at the equation LowlyPion wrote for you.
 
diazona said:
No you don't, you only have one, a. Look carefully at the equation LowlyPion wrote for you.

ok, i tried it, solved for a and I got .030 m/s^2 = a.

i typed that into my homework answers, and it said it was incorrect (we have multiple times to try it)

and when I do v=at,

3.8=(.030)t,

I get 125.3 seconds...seems a bit too much time. am i doing something wrong?
 
fireykitty said:
ok, i tried it, solved for a and I got .030 m/s^2 = a.

i typed that into my homework answers, and it said it was incorrect (we have multiple times to try it)

and when I do v=at,

3.8=(.030)t,

I get 125.3 seconds...seems a bit too much time. am i doing something wrong?

I trust you entered it as a = -.03 m/s2

It is decelerating after all.
 
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