# Help with this physics problem (acceleration in two dimensions).

• JetsetPyro
In summary, the hockey puck rebounds from a board and has an average acceleration of 7.5 N of W. To calculate the vector of the two velocities, you use sin(theta) and cos(theta). To find the modulus of a vector, you use |v_f| = p(\cos{(\theta)}\hat{i} + \sin{(\theta)}\hat{j}) and |a| = (|vf|-|vi|)/(/\t).
JetsetPyro
A hockey puck rebounds from a board as shown in Figure 16.

Figure 16 looks like this.

http://img174.imageshack.us/img174/2571/fig16vr6.png

The puck is in contact with the board for 2.5ms. Determine the average acceleration of the puck over the interval.

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So I know you have to use vector components. The answer says the avg. acceleration is 7.3x10^3m/s^2 [7.5 N of W] (degrees).

How do I solve it? =/

I found that the vector components for v1x and v2x to add up to 43.57, and the components of v1y and v1x to add up to (-1.87). However, I'm not getting the correct average acceleration value in the end.

Last edited by a moderator:
For the second vector the vector in the ight of the picture), what angle did you use?
The angle should be 180°-22° = 158°

I just tried it using 158 as the angle and I didn't get the correct answer given again.

I'm calculating the vector components using sin(theta)(Ay), sin(theta)By, cos(theta)Ax, and cos(theta)Bx of course.

I can't seem to find out what I'm doing wrong.

First you need to fix a coordinate system within which you'll work. Let the 'N' be our y-axis and 'E' our z-axis.

Now try to find the vector of the two velocities. For the first one, it is simply

$$\overrightarrow{v_i} = 26(\cos{(22)}\hat{i} - \sin{(22)}\hat{j})$$

it is important to note here that i have taken component on the 'y' axis to be negative. This is because, the direction of the 'y' component is downwards. Here is the diagram to resolve this vector:

http://img142.imageshack.us/img142/4125/resolvingonefc5.png

Now.. find the vector $v_f$ yourself. And then, remember that acceleration is given by:

$$\overrightarrow{a} = \frac{\overrightarrow{v_f} - \overrightarrow{v_i}}{\Delta t}$$

Last edited by a moderator:
|vf| = 21(sin(22)+cos(22))
|vf| = 27.337

|a| = (|vf|-|vi|)/(/\t)
|a| = (27.337-14.37)/0.0025s
|a|= 5186.8

Hmmm, did I do something wrong?

JetsetPyro said:
|vf| = 21(sin(22)+cos(22))
|vf| = 27.337

well.. for one.. you did many things wrong. First of all, you did not assign unit vectors to the $\overrightarrow{v_f}$. Also, there is a difference between $\overrightarrow{v_f}$ and $|v_f|$. Also, $|v_f|$ cannot be anything other than 21.

When finding the acceleration, you need to subtract vectors first and then take modulus. I suggest u refer your textbook and understand vectors.

I couldn't assign unit vectors to vf (with direction) as I don't know how to show that on this forum. =/

Also, we haven't been taught "modulus" or is that modules? I don't know what they are.

impulse

JetsetPyro said:
I couldn't assign unit vectors to vf (with direction) as I don't know how to show that on this forum. =/

Also, we haven't been taught "modulus" or is that modules? I don't know what they are.

To learn how to write in LaTeX, check this: https://www.physicsforums.com/showthread.php?t=8997

also.. u can show unit vectors using 'i' or 'j' like: vf = 21(cos(22)i + sin(22)j). Modulus of a vector is the 'magnitude' of the vector. If there is a vector:

$$\overrightarrow{l} = a\hat{i} + b\hat{j}$$

then,

$$|\overrightarrow{l}| = \sqrt{a^2 + b^2}$$

the modulus gives the length of the vector. For a velocity vector, the modulus is the magnitude of the vector, in the case of $v_f$, it is 21. For a vector in the form:

$$\overrightarrow{l} = p(\cos{(\theta)}\hat{i} + \sin{(\theta)}\hat{j})$$

the magnitude will always be 'p'. Also, even if u do it using impulse, it eventually will end up the same away. You'll assume a mass.. you'll find the change in momentum [which is nothing but the change in velocity multiplied by mass].. and then you'll divide it by time to find force and then reduce the 'm' to give u acceleration. Taking an extra 'm' will serve no purpose. So don't do it using impulse or collisions since only the acceleration is asked.

## 1. What is acceleration in two dimensions?

Acceleration in two dimensions is a measurement of the rate at which an object's velocity changes over time in two directions, typically represented by the x and y axes.

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

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. What are some common units of measurement for acceleration in two dimensions?

The most common units of measurement for acceleration in two dimensions are meters per second squared (m/s²) and feet per second squared (ft/s²).

## 4. How does acceleration in two dimensions differ from acceleration in one dimension?

Acceleration in one dimension only considers the change in velocity in one direction, while acceleration in two dimensions takes into account the change in velocity in two directions.

## 5. What are some real-world examples of acceleration in two dimensions?

Some real-world examples of acceleration in two dimensions include projectiles, such as a baseball being thrown or a rocket launching, and objects moving along a curved path, such as a car rounding a corner.

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