Constant acceleration of hockey puck

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Homework Help Overview

The problem involves a hockey puck that comes to rest after sliding a distance of 249 m, starting with an initial velocity of 4.5 m/s. The focus is on determining the constant acceleration of the puck during its motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion relevant to constant acceleration and express confusion about how to isolate time to find acceleration. One participant suggests using a different equation that relates initial and final velocities, displacement, and acceleration.

Discussion Status

Some participants have proposed an alternative equation to find acceleration, while others are still exploring the relationship between time, distance, and acceleration. There is no explicit consensus on the correct approach yet, but the dialogue is progressing with attempts to clarify the problem.

Contextual Notes

Participants note the absence of time in the initial equations, which complicates the solution process. The discussion reflects a focus on understanding the relationships between the variables involved in the problem.

anna_chem
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Homework Statement


A hockey puck sliding on a frozen lake comes to rest after traveling 249 m. If its initial velocity is 4.5 m/s, what is its acceleration if that acceleration is assumed constant?


Homework Equations


V(t)= v(initial) + at
d= v(initial)*t + .5at^2

The Attempt at a Solution


V(t)= 4.5 m/s + at or 249 m= 4.5 m/s*t + at^2

I'm confused as to where to go from here. How can I go about finding t, so that I can find a? Or am I trying to go about it wrong?
 
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anna_chem said:

Homework Statement


A hockey puck sliding on a frozen lake comes to rest after traveling 249 m. If its initial velocity is 4.5 m/s, what is its acceleration if that acceleration is assumed constant?


Homework Equations


V(t)= v(initial) + at
d= v(initial)*t + .5at^2

The Attempt at a Solution


V(t)= 4.5 m/s + at or 249 m= 4.5 m/s*t + at^2

I'm confused as to where to go from here. How can I go about finding t, so that I can find a? Or am I trying to go about it wrong?

Yes, that is one way, but you are missing time, so that equation can't be used.
We need an equation with initial,final velocity,displacement and acceleration

like this one

v2=u2+2ad
 
Thanks for the quick response. So, given that initial velocity is 4.5 m/s and final velocity is 0, then:
4.5^2=0+498a thus a=.0407 m/s^2
Does that look correct?
 
anna_chem said:
Thanks for the quick response. So, given that initial velocity is 4.5 m/s and final velocity is 0, then:
4.5^2=0+498a thus a=.0407 m/s^2
Does that look correct?

v2=u2+2ad

v=final velocity

u=initial velocity.
 

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