What is the acceleration of a hockey puck sliding on a frozen lake?

[JassC]

A hockey puck sliding on a frozen lake comes to rest after traveling 175m.
If it's initial velocity is 1.0m/s, what is its acceleration if that acceleration is assumed constatnt?
Answer in units of m/s^2

I first divided 175/1.9 to get 92.11s. I then put 1.9/92.11 to get 0.02063. But it's wrong.

Any ideas?

Thanks.

 

[d_leet]

A hockey puck sliding on a frozen lake comes to rest after traveling 175m.
If it's initial velocity is 1.0m/s, what is its acceleration if that acceleration is assumed constatnt?
Answer in units of m/s^2

I first divided 175/1.9 to get 92.11s. I then put 1.9/92.11 to get 0.02063. But it's wrong.

Any ideas?

Thanks.

Where did 1.9 come from for the first thing. For the second this object is accelerating so what equations do you know for accelerated motion that take account of the information you have?
-The distance it travels
-It's initial velocity
-It's final velocity

 

[Doc Al]

Not sure what you're doing here. What's the 1.9 signify?

Start by writing the kinematic relationship between velocity, acceleration, and distance.

 

[JassC]

Sorry...the initial velocity is suppose to be 1.9m/s...my bad.

 

[Doc Al]

In that case, realize that the speed is not constant. To find the time as you attempted, you must use the average speed. (What's the average speed?)

 

[JassC]

1.9/2 = 0.95 (Average speed)

175/.95 = 184.2105s (Time took for puck to stop)

I'm stuck here now.

 

[Doc Al]

Now apply the definition of acceleration.

 

[JassC]

0.95/184.2105 = 0.005157

1.9 + (0.005157)(184.2105) = 2.85 m/s^2

Hopefully I've gotten it right.

 

[Doc Al]

Again, I'm not sure what you are doing with these equations. What's the definition of acceleration? That's all you need.

 

[JassC]

Umm the definition? The rate of change? or do you mean speed over time...

 

[Kurdt]

the velocity was 1.9m/s and it changed to 0m/s in a time you calculated. the correct definition is the rate of change of velocity with time.

Does this help?

 

[Doc Al]

$$a = \Delta v / \Delta t$$

 

[JassC]

Is the answer 0.005157? I'm so confused right now.

 

[Doc Al]

Just apply the definition. What's the change in velocity? (You already calculated the time.)

 

[JassC]

I did use the definition...speed over time and which is 0.95/184.2105 = 0.005157

I have no clue how to find the change in velocity...Sorry, I'm new to phyiscs. Just started learning...

 

[Doc Al]

The definition is change in speed over time, not just speed over time. To find the change in anything, take the final value and subtract the initial value. What's the final speed? What's the initial speed?

Don't confuse the average speed, which you had to use to find the time, with the initial or final speed, which is what you need to find the acceleration.

 

[JassC]

I'm so lost...I don't know how to find the final value...

My best guess is 0.95 final speed(since that's what I got from 175/184.2105) - 1.9 initial speed...which gives you -0.95.

 

[Doc Al]

I don't know how to find the final value...

Sure you do, if you read the problem carefully. I'll quote: "A hockey puck sliding on a frozen lake comes to rest..."

My best guess is ...

No guessing!

 

[JassC]

well then its 0...

 

[Doc Al]

Of course. The final velocity is zero. So what's the change in velocity?

 

[JassC]

-1.9 is the change in velocity...

-1.9/184.2105 = -0.01031?

 

[Doc Al]

Right!

Be sure to:
- label your calculations: "acceleration = ..."
- use appropriate units: m/s^2
- round off the answer to a reasonable number of significant figures

 

[JassC]

Yeah...but the question's not over :(

How long is it in motion? Answer in units of s.

I don't know the equation of motion.

 

[Doc Al]

Hint: You already solved that question!

 

[JassC]

Ohhh right, thanks a lot for your help =)

 

[JassC]

Just when I thought I had the last part all figured out...I don't.

What is its speed after traveling 125m? Answer in units of m/s.

I thought it would just be replacing the 175m with 125m. And then just doing the same steps in Part 1 to get the answer. I was wrong lol.

 

[Kurdt]

Calculate how long it takes to reach 125m the same as you have done for 175m. You know the acceleration of the puck so you can find the change of speed in this time and add it to the initial velocity.

 

[Tensaiga]

Hello, I'm new to physics too, but i think i know how to do this... don't you just use the formula:
(velocity final)^2 - (Velocity initial)^2 = (acceleration)*(displacement)
This formula: $$v^2 = v_0^2 + 2 a \Delta x$$

Some Thoughts:
i don't know if you learned this or not. But since you didn't APPLY any force to it, it's totally stopped by friction, which means that the acceleration is in opposite of the puck's direction of moition. So the acceleration is negative. No need to convert the units, i think they are in the right format.

Now, just plug in the numbers, velocity finial is 0m/s, it comes to rest after 175m. velocity inital is 1m/s, and the displacement is 175.
SO:
(0m/s)^2 - (1m/s)^2 = 2(a)(175m)
a = -(1m/s)^2/(2)(175m)
a = -0.002857142m/s^2

well that's what i got, I'm a newbie too. lol.

PS: you don't need the time variable in this equation saves you time.

AND how long it has been in motion:
just use : $$v = v_0 + a t$$
velocity finial = velocity inital + (accerlation)*(time)
(0m/s) = (1m/s) + (-0.002857142m/s^2)*(time)
time = - (1m/s)/(-0.002857142m/s^2)
time = 350s.

you could check this answer to see if it gives you a distance of 175m.

PS: i checked and it does.lol.
I used this formula : $$x = x_0 + v_0 t + (1/2) a t^2$$

Oh by the way, the velocity after 125m, just use my first formula,$$v^2 = v_0^2 + 2 a \Delta x$$, expect this time the displacement is 125m, instead of 175. (HINT:You already known the accerlation this time.)

 

[borisleprof]

When the acceleration is constant, remember these equations:
<v> = (vf +vi)/2 = mean velocity.
If t = time required to stop,
distance required to stop = d = <v>t
vf = final velocity
vi = initial velocity
Now y ou must remember the definition of acceleration
acceleration = a= (? - ?)/?

replace the questions marks

 

[borisleprof]

Remember, think physics first, whenever you are new to physics
Don't just play with formulas.
Use basic concepts.
Know how to calculate distance.
The only few thing you must really know is this
distance = mean velocity * time
acceleration = (final velocity- initial velocity)/(time it takes to pass from initial velocity to final velocity)
And when the acceleration is constant you must know how to calculate the mean velocity.
<v> = (final velocity + initial velocity)/2

bye

 

[daniiiiiii]

A hockey puck sliding on a frozen lake comes to rest after traveling 175m.
If it's initial velocity is 1.0m/s, what is its acceleration if that acceleration is assumed constatnt?
Answer in units of m/s^2

I first divided 175/1.9 to get 92.11s. I then put 1.9/92.11 to get 0.02063. But it's wrong.

hi,if body moves linear then 2(acceleration)(distance)=(final velocity)^2 - (initial velocity)^2 fomulae can work. so d=175m and intial velocity=1.0m/s^2 and final velocity=0m/s^2