Find Acceleration with only Initial Velocity and Final position?

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Homework Help Overview

The problem involves a hockey puck that comes to rest after sliding a distance of 238 m, starting from an initial velocity of 3.8 m/s. The objective is to determine the puck's acceleration, assuming it is constant, and to explore related questions about the time of motion and speed at a distance of 180 m.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations, particularly focusing on the equation v² = (vo)² + 2a(x - xo). There is uncertainty regarding the final velocity and the role of time in the calculations. Some participants express confusion about how to isolate variables and solve for acceleration without knowing the time.

Discussion Status

Participants are actively engaging with the problem, offering hints and discussing the implications of the phrase "comes to rest." Some have attempted calculations for acceleration and time but have encountered discrepancies in their results. There is no explicit consensus on the correct approach or solution yet, as participants continue to explore different interpretations and calculations.

Contextual Notes

Participants note that the problem may involve assumptions about constant acceleration and the implications of deceleration, as well as the potential for multiple attempts in the homework system leading to confusion over correctness.

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 + volt +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.
 
Last edited:
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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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