Problems with kinematics equations

[Byrgg]

I have a few homework questions involving some kinematics equations, vectors, etc. I was hoping I could some help with them.

My first question:

An arrow strikes a target in an archery tournament. The arrow undergoes an average acceleration of 1.37 x 10^3 m/s^2 [W] in 3.12 x 10^-2 s, then stops. Determine the velocity of the arrow when it hits the target.

I tried this, and my answer didn't match the one in the book, I started with this:

v_2 = v_1 + at

And I wasn't sure this was even the right equation, or how to stick the numbers in. Time and acceleration were easy to substitute in, but then I didn't know what to do with the two velocities, one of which, I'm assuming is 0 m/s.


Another problem I have reads as follows:

Starting with the defining equation for constant acceleration and the equation for displacemnt in terms of average velocity, derive the constant aceleration equation

a) from which final velocity has been eliminated
b) from which initial velocity has been eliminated

I got about as far as picking the equations that it says to start with:

a = (v_2 - V_1)/t and v = d/t

I''m not sure what to do next, please help me figure this out.


Another one went as follows:

A watercraft with an initial velocity of 6.4 m/s [E] undergoes an average acceleration of 2.0 m/s^2 for 2.5 s. What is the final velocity of the watercraft?

I got confused by this, because the acceleration is perpendicular to the velocity, and so I wasn't really sure how to start, any help would appreciated.


Lastly(I hope), is this one:

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.

Figure 16 shows the puck hitting the board with a velocity of 26 m/s 22 degrees from the board, and then rebounding off at the same angle with a velocity of 21 m/s.

I really wasn't sure how to go about starting this, so any guidance here would be appreciated.

 

[Astronuc]

Looking at the first problem

v_2 = v_1 + at

one has to be careful when using formulas. One must distinguish between absolut time and duration (or difference in time).

Let v₂ = v (t₂) and v₁ = v (t₁), then v₂ = v₁ + a(t₂-t₁), where a is the average acceleration between t₂ and t₁. Let 1 reference the initial state and 2 reference the final state.

When the arrow stops, v₂ = 0, so the previous equation becomes, v₁ + a(t₂-t₁) = 0

Then v₁ = -a(t₂-t₁)

Now the problem states "The arrow undergoes an average acceleration of 1.37 x 10^3 m/s^2 [W] in 3.12 x 10^-2 s, then stops."

So, (t₂-t₁) = 0.0312 s, and the acceleration should be a deceleration, or acceleration of -1379 m/s², so

v₁ =

-----------------------------------------------------------------

See if this helps.
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1

 

[Byrgg]

43 m/s [E], right? My textbook says 42.8, not sure how that came about though.

 

[Astronuc]

I got about 43 m/s. I see the mistake on my part, I wrote 1379, instead of 1370 m/s². Using 1370, velocity is 42.7 m/s. My apologies for the error.

I take [E] and [W] refer to east and west, respectively.

 

[Byrgg]

Ok, thanks for the help with that question. Would you be able to help me with the others as well?

 

[Astronuc]

We are pleased to assist, but please show some work on each problem, e.g. write the formulas and substitute in the values.

 

[Byrgg]

Subsituting is the problem, for my second question. I wasn't sure how to substitute everything in, since it's a derivation, and a little complicated, I think. I'm pretty sure my forumlas were right, I'm just not sure what to do with them.

For my third question, I tried to separate the acceleration into x and y components, and since it's direction South, it only has a y component. This means that the acceleration caused by it is only in the South direction, so I'm guessing that also means that the velocity is change is also in the South direction. The problem is, I don't know how to show the calculate the change in velocity, I was thinking of this:

V_2 = V_1 + a(t_2 - t_1)

And then using zero as the initial velcoity, but that didn't make sense to me, for some reason. The acceleration and time were easy to substitute in, but I'm not sure if this is right, or how to explain it.

As for my third question, I guess 26 m/s [22 degrees S of E] would be the initial velocity, whereas 21 m/s [22 degrees N of E] would be the final velocity, and then the change in time is given to be 2.5 ms. The velocities are in different directions though, so would I have to break them up into their x and y components in order to work with them?

Sorry if it's not much, but this is about all I can come up with, at least I think it's a little better than before.

 

[Doc Al]

Another problem I have reads as follows:

Starting with the defining equation for constant acceleration and the equation for displacemnt in terms of average velocity, derive the constant aceleration equation

a) from which final velocity has been eliminated
b) from which initial velocity has been eliminated

I got about as far as picking the equations that it says to start with:

a = (v_2 - V_1)/t and v = d/t

I''m not sure what to do next, please help me figure this out.

The equation for acceleration is fine. Express the second equation in terms of initial and final velocities. (In that equation, "v" is the average velocity.)

Once you've done that you'll have two equations in terms of V_1 and V_2. To eliminate a variable, solve for it in one equation, then substitute it into the other.

 

[Doc Al]

For my third question, I tried to separate the acceleration into x and y components, and since it's direction South, it only has a y component. This means that the acceleration caused by it is only in the South direction, so I'm guessing that also means that the velocity is change is also in the South direction. The problem is, I don't know how to show the calculate the change in velocity, I was thinking of this:

V_2 = V_1 + a(t_2 - t_1)

And then using zero as the initial velcoity, but that didn't make sense to me, for some reason. The acceleration and time were easy to substitute in, but I'm not sure if this is right, or how to explain it.

You are exactly on the right track. Hints: Does the x-component of velocity change? What is the final y-component of velocity? (What's the initial y-component of velocity?)

As for my third question, I guess 26 m/s [22 degrees S of E] would be the initial velocity, whereas 21 m/s [22 degrees N of E] would be the final velocity, and then the change in time is given to be 2.5 ms. The velocities are in different directions though, so would I have to break them up into their x and y components in order to work with them?

It's easiest if you treat the x and y components separately. Find the x and y components of the initial and final velocities. (You don't have to break them into components, if you are comfortable finding the change between initial and final velocity vectors.)

 

[Byrgg]

I have a question about the arrow problem, how do I find out the direction of the velocity? The book says East, but my calculations say West, did I do something wrong?

 

[Doc Al]

I have a question about the arrow problem, how do I find out the direction of the velocity? The book says East, but my calculations say West, did I do something wrong?

Show what you did. I would call the positive x-axis to be east, in writing your equations.

 

[Byrgg]

Yeah, that's what I did, but the acceleration is west, and the arrow doesn't seem to have an initial velocity.

 

[Doc Al]

...and the arrow doesn't seem to have an initial velocity.

Seriously, does that make any sense at all?

In your first post you gave the proper equation. Show how you used it.

 

[Byrgg]

I meant that the arrow's initial velocity is 0 m/s, not that it was non exsistant, come to think of it, that really doesn't make any sense, since the final velocity is 0 m/s, then the overall acceleration would be 0.

Woops, now I think I know what I did wrong, the final velocity is 0 m/s. so when you rearrange the equation the accelaration becomes negative, and since it's already in the negative direction, it becomes positive, resulting in a positive velocity.

 

[Byrgg]

I just figure out the derivations, I think I'm doing something wrong. I start with these, right?

a = (v_2 - V_1)/t and d = 1/2(v_2 - v_1)(t)

Things never seem to work out when I try to substitute and rearrange them, could someone help me please?

 

[JMuse]

When you average numbers, you add them, not subtract them.

 

[Byrgg]

Oh, I forgot about that, it's a bad habit since I'm usually calculating the difference. I'll see if I can figure it out now, thanks for reminding me about that, though I feel like an idiot now, many of my problems on individual homework questions come from carelss mistakes like that.

 

[Byrgg]

I think I got it right, finally:

a = (v_2 - V_1)/t and d = 1/2(v_2 + v_1)(t)

2d = (v_2 + v_1)(t)
2d/t = v_2 + v_1
2d/t - v_1 = v_2

a = (v_2 - v_1)/t
a = ((2d/t - v_1) - v_1)/t
at = (2d/t - v_1) - v_ 1
at + 2v_1 = 2d/t
at^2 + 2v_1t = 2d
d = v_1t + 1/2at^2

Now I just have to do the other one, and then the rest of the questions.

 

[Byrgg]

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.

Figure 16 shows the puck hitting the board with a velocity of 26 m/s 22 degrees from the board, and then rebounding off at the same angle with a velocity of 21 m/s.

This is one my earlier questions, the only I have yet to solve, well, there's another one in my book, but I haven't looked at it yet. I was wondering if someone could help me a bit with this. I can probably calculate the resultant vector using the cosine law, but something doesn't make any sense to me. The only time given is the time during which the puck is in contact with the board. The entire trip is more than just the time when the puck is in contact with the board, so this doesn't really make any sense, someone please help me figure this out.

 

[Byrgg]

Someone please help.

 

[Doc Al]

The only time given is the time during which the puck is in contact with the board. The entire trip is more than just the time when the puck is in contact with the board, so this doesn't really make any sense, someone please help me figure this out..

All they are interested in is the average acceleration of the puck while it is in contact with the board. The other part of its motion is irrelevant.

(Just like with the arrow problem: The acceleration given was the arrow's average acceleration during its interaction with the target, not during its travel from the bow to the target.)

 

[Byrgg]

Ok, then I would just use the velocities provided for the initial and final velocities? I guess that makes sense, thanks for the help.

 

[Byrgg]

Is there anyone here who can help me right now? I couldn't get the last question, and the next question is confusing me, as well.

 

[Byrgg]

I'll post my other question, but I still need help with the last one, here's the new one:

A passenger in a hot-air balloon throws a ball with an initial unknown velocity. The ball accelerates at 9.8m/s^2 [down] for 2.0s, at which time it's instantaneous velocity is 24m/s [45 degrees below the horizontal. Determine the ball's initial velocity.

I'm guessing you have to break up the final velocity into it's x and y components to start, but I'm not entirely sure, could someone give me a hand with this, as well as my other question?

 

[Doc Al]

A passenger in a hot-air balloon ...

I'm guessing you have to break up the final velocity into it's x and y components to start

Yes, break it into components and deal with each separately.

When you ask about multiple problems in the same thread, things get confusing fast. At least number them. And stick to one problem at a time in any given thread until that problem is resolved. If you want to discuss several problems at once, use a new thread for each.

 

[Byrgg]

Ok, sorry about dealing with so many things at once, could I get some help with the hockey puck question first?

 

[Doc Al]

could I get some help with the hockey puck question first?

You asked a question about that problem in post #19, which I thought I answered in post #21. Do you have another question? Where are you stuck? What have you done so far?

 

[Byrgg]

Well, my final answer(for the hockey puck question) ended up wrong(apparently), I didn't really know what I was doing for the last part of it anyway, so it's to be expected. My work is as follows:

$2.5ms = 2.5 x 10^-3 s$

(I just realized that I think my intial conversion there was wrong, and is now corrected, but I'll see how everything works out with the new number)

(cosine law to get the resultant vector for velocity)

$v_t^2 = v_1^2 + v_2^2 - 2(v_1)(v_2)cos136$
$v_t = \sqrt{v_1^2 + v_2^2 - 2(v_1)(v_2)cos136} = \sqrt{26^2 + 21^2 - 2(26)(21)cos136} = 18$

(I may have made a mistake in there, but I'm pretty sure I did it right)

(the overall acceleration I'm not sure how to calculate, because I'm not sure what to use as the initial velocity)

(calculating the angle of the velocity)

$sin\theta/21 = sin136/18.2 \theta = 53$

$53 - 22 = 31$

By my calculations, the angle above the horizontal should be 31degrees, whereas the book says 10 degrees above the horizontal, now I know the books sometimes have mistakes in them, but I really think I'm doing something wrong here. Along with getting the wrong angle(apparently), I don't know how to go about calculating the average acceleration, because I'm not sure what to use for the initial velocity. That work there is about all I've come up with, and I'm sure it's wrong in some way or another, so some help would be greatly appreciated.

 

[Byrgg]

Someone please help.

 

[Doc Al]

(cosine law to get the resultant vector for velocity)

You are trying to find the change in velocity:
$$\vec{v}_f - \vec{v}_i$$

$v_t^2 = v_1^2 + v_2^2 - 2(v_1)(v_2)cos136$
$v_t = \sqrt{v_1^2 + v_2^2 - 2(v_1)(v_2)cos136} = \sqrt{26^2 + 21^2 - 2(26)(21)cos136} = 18$

The angle between the sides is not 136, but 44. (But the resulting magnitude is the same.) Draw a picture to clarify the angles involved. I assume "22 degrees to the board" means that the velocity makes a 22 degree angle with the surface of the board.

(the overall acceleration I'm not sure how to calculate, because I'm not sure what to use as the initial velocity)

You are given the initial and final velocities.

(calculating the angle of the velocity)

Until we see a picture it will be hard to evaluate your angles.

 

[Byrgg]

The angle between the sides is not 136, but 44. (But the resulting magnitude is the same.) Draw a picture to clarify the angles involved. I assume "22 degrees to the board" means that the velocity makes a 22 degree angle with the surface of the board.

How is it 44? The velocities make a v shape with the board, 22 degrees to the board each way. Knowing that the board is 180 degrees, the angle between them should be 180 - (22 + 22), shouldn't it?

 

[Doc Al]

How is it 44? The velocities make a v shape with the board, 22 degrees to the board each way. Knowing that the board is 180 degrees, the angle between them should be 180 - (22 + 22), shouldn't it?

That's the angle between those velocity vectors, but it's not the angle in the triangle that defines the difference between those two vectors:
$$\vec{v}_f - \vec{v}_i$$

Again, draw a picture.

 

[Byrgg]

Ok, I'll draw a picture later at some point, I'm at school right now, so I can't really draw it now, but I'll make one at home, and show exactly how the diagram is set up. Hopefully then you'll be able to help me get a better understanding of this. How long will it take to get an attachment approved, approximately? Or could I just use an image hosting site, and post a link to the diagram?

 

[Doc Al]

How long will it take to get an attachment approved, approximately?

If someone's paying attention, about 2 seconds.

Or could I just use an image hosting site, and post a link to the diagram?

That's OK too.

Be sure to label everything in the diagram clearly.