Is this correct without a direction in metres?

[barthayn]

Homework Statement

A rocket-powered hockey puck moves on a horizontal frictionless table. Figure EX4.8 at the top of the next column shows graphs of vx and vy, the x- and y-components of the puck's velocity. The puck starts at the origin.

a. In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis.
b. How far from the origin is the puck at t = 5s?

The graph of the X-velocity is v=8t cm/s and the Y-velocity is V=30cm/s

Homework Equations

c^2 = a^2 + b^2
d = vt
m= dy/dx

The Attempt at a Solution

a. vx = 16 cm/s. vy = 30cm/s. θ = tan (30/16)^-1 = 62^o. Therefore the puck is traveling in the direction of 62^0 from the x-axis.

b. vx = 8* 5 = 40cm/s
dx = 40*5 = 200cm

dy= 30*5 = 150cm

c = (4000+2250)^0.5 = 250cm

Therefore, the puck is 250 cm from the origin.

My question are these:

For a I do not need to add the displacement of it, correct? Just the angle is good enough right? In high school there was never a question that did not involve the magnitude of the displacement.

Secondly. Is the position traveled from the origin correct as well? I am asking this because I am a bit confused because the position function for x-direction is 4x^2. It shouldn't matter if it is linear or not correct. I can still do it the way I solved it correct? As well for b, I do not need to include the angle because it is asking for the magnitude of the displacement, correct?

 

[NascentOxygen]

Your (a) looks correct.

b. vx = 8* 5 = 40cm/s
dx = 40*5 = 200cm

200 cm is the distance it would have covered were it traveling at a constant speed of 40cm/s since the start. It has been traveling at less than this all the while, so your answer cannot be correct.

You need to find the displacement (from the origin) at time t=5 secs. The magnitude of this gives you the distance from the origin at t=5. Displacement is found as the area under the v-t graph. You have the graph, so you can work out the area under it.

Area has units. What units will the area under your v-t graph have?

 

[barthayn]

Your (a) looks correct.



200 cm is the distance it would have covered were it traveling at a constant speed of 40cm/s since the start. It has been traveling at less than this all the while, so your answer cannot be correct.

You need to find the displacement (from the origin) at time t=5 secs. The magnitude of this gives you the distance from the origin at t=5. Displacement is found as the area under the v-t graph. You have the graph, so you can work out the area under it.

Area has units. What units will the area under your v-t graph have?

Thanks, I forgot to half the area found because it was a triangle. To prove me correct I took the integral. I got 100 cm from x. So the final answer is 180.27cm from the origin, correct?

 

[NascentOxygen]

My answer is 180.27cm also.

 

[rwisz]

As a scientist, it is important to be precise and clear in your response. Here are some suggestions for how you could improve your response:

- For part a, it would be helpful to specify that the angle is being measured counterclockwise from the positive x-axis, as this is the standard convention.
- It is not necessary to add the displacement for part a, as you correctly stated. However, it would be helpful to mention that the magnitude of the velocity is not needed for this part either, as the angle can be determined solely from the components of the velocity.
- For part b, the equation you used to find the displacement is not correct. The displacement should be calculated using the average velocity, which is (vx+vy)/2. Also, it would be helpful to mention that the angle is not needed for this part, as it is asking for the distance from the origin, not the displacement.
- In your solution for part b, you used the position function for the x-direction, but this is not relevant to the problem. The position of the puck can be found by multiplying the average velocity by the time, as you correctly stated.
- It is important to use the correct units in your calculations and final answer. In this case, the velocity components are given in cm/s, so the displacement and distance should also be given in cm.
- Lastly, it would be helpful to provide a clear and concise conclusion or summary of your answer, rather than just stating that it is correct. This will help to reinforce your understanding of the problem and provide a more complete response.

Overall, your approach to the problem is correct, but it would be beneficial to clarify and expand on some of the steps and concepts involved.