# Acceleration in two dimensions. Hockey puck question

• TimelessTime
In summary: Another way to see this is to note that the horizontal speed changes by only 4.6 m/s, while the vertical speed changes by 17.6 m/s (going from 21 to -17.6). So the change in vertical direction is quite large, and the change in horizontal is quite small. The overall vector points in the average direction of these two changes, which is close to 75 degrees N of W.So your answer is right. I think we can put this thread to bed now. Good night!In summary, the problem is to determine the average acceleration of a hockey puck rebounding from a board, with a contact time of 2.5 ms. After several attempts and
TimelessTime
Problem: A hockey puck rebounds from a board as shown in figure 16. The puck is in contact with the board for 2.5 ms. Determine the average acceleration of the puck over the interval.

t = 0.0025 s |vi|= 26 m/s |vf| = 21 m/s aav = ? *v and a are vectors*

7.3 x 10^3 m/s^2[7.5 N of W]. In my opinion my closest attempt was that I broke the initial vector and final vector into components, found the change in vx to equal 4.6 m/s and the change in vy to equal -1.9 and then found the change in v to equal 5. my final answer and direction were completely wrong. Can someone help me with this awful question?

#### Attachments

• Figure 16.png
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Welcome to PF TimelessTime!

Don't worry, this question is quite straightforward, and you already have the right method! The average acceleration is given by change in velocity divided by change in time:

a = Δv/Δt

The thing that needs to be done is finding the change in velocity Δv, in each direction, using those to calculate separate horizontal and vertical components of the acceleration, and then combining these into the overall vector. Could you please post your work (the actual calculations you used to find the change in horizontal and vertical speeds), so that we can track down where you went wrong?

Also: working through some of the numbers, here is another hint: pay very close attention to the signs of your components. Are they in the same direction both before and after the rebound? This affects your computation of the Δv's

These were my calculations:
|Vx| = |Vfx| + (-|Vix|)
= 21 m/s(sin68) - 26 m/s(cos22)
= -4.6 m/2 oops, i got -4.6 but on the next page i left out the negative, not sure if that is the real problem though

|Vy| = |Vfy| + (-|Viy|)
= 21 m/s(cos68) - 26 m/s(sin22)
= -1.9
|V| = sqrt((-1.9)^2 + (-4.6)^2)
= 5.0 m/s
|aav| = 5/0.0025 != 7.3 x 10^3

TimelessTime said:
These were my calculations:
|Vx| = |Vfx| + (-|Vix|)
= 21 m/s(sin68) - 26 m/s(cos22)
= -4.6 m/2 oops, i got -4.6 but on the next page i left out the negative, not sure if that is the real problem though

|Vy| = |Vfy| + (-|Viy|)
= 21 m/s(cos68) - 26 m/s(sin22)
= -1.9
|V| = sqrt((-1.9)^2 + (-4.6)^2)
= 5.0 m/s
|aav| = 5/0.0025 != 7.3 x 10^3

One of your two components of Δv is wrong here. I'll reiterate my point from post #3. Is the puck going in the same direction (along the east-west axis) after the rebound? What about along the north-south axis? Same direction as before, or opposite?

so what the problem is is that for the x components the should both be positive and for the y components it is a subtraction?

There's another problem as well. Don't compute the final acceleration from the magnitude of the velocity. Remember, acceleration is a VECTOR. So you should calculate the individual components of a first, and then combine them:

ax = Δvx/Δt

ay = Δvy/Δt

a = (ax2 + ay2)1/2

alright, ill try solving it again and post if there's anymore problems. Thanks for the help.

TimelessTime said:
so what the problem is is that for the x components the should both be positive and for the y components it is a subtraction?

You tell me. Just be systematic about it. Let's call north the positive vertical direction, and south the negative vertical direction. So if the vertical component of the vector is northward, it should be positive. If it is southward, it is negative.

Likewise for the horizontal components. Call east + and west -

Initially, is the puck going east or west? What about finally? So what is the sign of the horizontal component before and the one after?

Initially, is the puck going north or south? What about finally? So what is the sign of the vertical component before, and the one after?

These questions I've been asking you are not rhetorical, I intended for you to actually answer them. I'm trying to walk you through the reasoning.

Ok, so I found the average acceleration to be 7.3 x 10^3 however, the direction I got was 75degrees N of W. Here's what I did:

|Vx| = 21 m/s(cos22) - 26 m/s(cos22)
= -4.64 m/s

|Vy| = 21 m/s(sin22) + 26 m/s(sin22)
= 17.6 m/s

aav,x = -4.64/0.0025 = -1856 m/s^2

aav,y = 17.6/0.0025 = 7040 m/s^2

Then, aav = sqrt(7040^2 + (-1856)^2) = 7.3 x 10^3

for the direction I found theta in the pic I attached and it ended up being 75NofW instead of 7.5NofW

#### Attachments

• average hockey puck.png
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7.5 degrees N of W must be a misprint. 75 is right. It must be fairly close to perpendicular to the wall.

## 1. What is acceleration in two dimensions?

Acceleration in two dimensions refers to the rate of change of velocity of an object in two different directions. It takes into account both the magnitude and direction of the change in velocity.

## 2. How is acceleration calculated in two dimensions?

Acceleration in two dimensions can be calculated using the formula a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 3. How does acceleration in two dimensions affect a hockey puck?

In the context of a hockey puck, acceleration in two dimensions refers to the changes in its speed and direction as it moves across the ice. This can be affected by external forces such as friction, gravity, and the player's stick.

## 4. What factors can influence the acceleration of a hockey puck?

The acceleration of a hockey puck can be influenced by several factors, including the force applied by the player's stick, friction between the puck and the ice, air resistance, and the angle at which the stick hits the puck.

## 5. How is acceleration in two dimensions important in understanding the motion of a hockey puck?

Acceleration in two dimensions is crucial in understanding the motion of a hockey puck as it allows us to analyze how the puck's speed and direction change over time. This information can help players make strategic decisions and improve their performance on the ice.

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