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Offline BFM_Xtr3me

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physics/maths question
« on: March 14, 2011, 01:13:54 AM »
hey, i got 2 questions from my engineering physics class and im not sure on how to do them?
here they are

When startled, an armadillo will leap upward. Suppose it rises 0.544m in the first 0.200 seconds

a) What is its initial speed as it leaves the ground?
b) What is it's speed at a height of 0.544m?
c) How much higher does it go?



the second question involved a graph so i scanned it and uploaded it
























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Offline BFM_dStruct

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Re: physics/maths question
« Reply #1 on: March 14, 2011, 01:25:19 AM »
I did Physics a while back, but isn't the equation something like

Speed = Distance over Time?

So 1 a Would be

Speed = 0.544m / .200

= 2.72 m per second


Offline BFM_Xtr3me

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Re: physics/maths question
« Reply #2 on: March 14, 2011, 04:27:14 AM »
nah, it has to do with the constant acceleration formulas,
since gravity is involved, speed will never be the same (unless a constant force is added),
and i think i just answered me own question xd

i need a good physics expert to tell me whether this is right or wrong
using the formula
s=ut + 1/2(a*t^2)
where s=0.544m, t=0.2secs, a=-9.8m/s^2 and u=initial velocity
the only thing im unsure of is the acceleration


i also still need help with the 2nd question


























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Offline jim360

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Re: physics/maths question
« Reply #3 on: March 14, 2011, 04:41:25 AM »
I'm not going to work through the calculations, but the way to approach this problem is based on what is known over here as the "SUVAT" equations. These tell you all about any movement where the acceleration is a constant.

The equations are:

a=constant
v=u+at
s=ut+(1/2)*at2
v2=u2+2as

Here are the meanings of the letters:

a=acceleration
u = initial velocity (at time t=0) (velocity is the slightly more accurate word for speed)
t = time
s = distance moved from the starting point
v = velocity at the time t

How to use these? What you have to do is look at the question, work out what you know and what you're looking for, then find which equation is the most suitable.

In your first example:

When startled, an armadillo will leap upward. Suppose it rises 0.544m in the first 0.200 seconds.

a) What is its initial speed as it leaves the ground?

 
We want to find its initial speed, and we are given the distance moved, the time taken, and we also know the acceleration - even though it's not written down. THis is because we can take it as given that this is happening on earth, so we use the value for acceleration due to gravity, g = -9.81 metres per second per second (don't forget the minus sign/ negative sign!)

so that means we know t, a and s and want to find u. We look out our equations and find that the equation we are looking for is:

s = ut + (1/2)at2

Now, to find u from that, we need to do some rearranging. After a couple of quick steps, you'll see that:

(s - (1/2)at2)/t = u

Then we replace t with 0.2, s with 0.544 and a with 9.81, and solve this. Importantly this will lead to a different answer from dStruct's, because we've taken into account the effect of gravity. Why? Well, assuming dStruct was right, then in fact the armadillo would carry on rising for ever! This isn't going to happen - "What goes up must come down" - so his answer is missing something.

For the second part:

b) What is its speed at a height of 0.544m?

Now we are wanting to find v, and we know s and a, and also t (which is still just 0.2 seconds), and we've just worked out u from before. So we could use either:

v = u + at

or:

v2=u2+2as

Or indeed both, to check that they agree (if they don't then you've made a mistake in the calculation somewhere).

What about the third part:

c) How much higher does it go?

This is harder because we don't know when it will reach its maximum height ( so we don't know t). What do we know? We still know the acceleration and the initial velocity and we want to find the distance moved, so that means we want an equation with u, a and s. Now here's the nice thing - we also know v, the speed it's moving at when it reaches its maximum height, which is 0!

Why can I say that? This is why calling speed "velocity" comes in handy, and why I said that the acceleration was -9.81 and not 9.81.

Speed is a number that tells you how fast you're moving. Velocity also tells you how fast you're moving, but it tells you something else - what direction you're moving in. In this question the armadillo moves up, then down. So what we'll do is call the upward motion positive, and the downward motion negative. This is why the acceleration due to gravity is -9.81, because it's dragging the armadillo downward.

Just before the armadillo reaches its maximum height, it's heading up, and just afterwards it's heading down again. So before its peak it's travelling at a velocity greater than zero, and afterwards it's travelling at a velocity less than zero. So, in between, it's travelling at a velocity of zero.

Returning to the question, we have the handy equation:

v2=u2+2as

We now know that v = 0 at the maximum height, so rearranging gives:

s = u2/2a

where now we use a = 9.81 metres per second per second (no minus sign)*.

Now this number isn't quite the right answer, because the question was "How much higher [than 0.544m] does it go?", so we could just subtract 0.544m from whatever our answer is. Or, to save the bother, we could just replace the u in this equation by the answer we got for part (b). This will give the same answer, try it and see!

This should complete the problem.

*Technically a is still -9.81, but when rearranging the equation another minus sign cropped up, so they've just cancelled.



For the second problem there's a bit of deduction needed because the graph they have presented actually is describing two different parts of the motion at once.

The same equations that we were using last time still apply, because again the acceleration is a constant -g = -9.81 metres per second per second (we'll want the minus sign again!).

However, things are very slightly complicated by the fact that the ball isn't travelling up, then down, but rather follows a curved path looking something like this:



As the ball travels its velocity is not only changing in size, but also in direction, as shown in that picture. In your graph only the change in value of the velocity has been included. To take account of the direction change we can consider two directions: up/ down (which we will call y) and left/ right (which we'll call x).

These two directions are independent of each other! This means that we can consider two separate sets of equations: the up/ down motion and the left/ right motion.

Start by thinking about the left/ right motion, with these equations:

a= constant
v=u +at
x=ut + (1/2)*at2
v2=u2 + 2ax

However, gravity only acts downward and has therefore got no effect on the left/ right motion! In fact there's nothing affecting the left/ right motion, so acceleration is just 0. Our equations become:

v=u
x=ut

So if we know the horizontal speed at any point in the flight of the ball, we know it everywhere else in that flight. We consider the speed of the ball at its maximum height and (if you believe my diagram) this tells us that u = 19 m/s horizontally.

What about the vertical motion? This is affected by gravity, so we have:


vy=uy - gt
y=uyt - (1/2)*gt2
vy2=uy2 - 2gy

Here vy and uy just mean the velocity and initial velocity in the y direction. The problem here is that we still haven't found out what u is. It's not 31, nor is it 19, and it's not even 11 (you might think it's 31-19) so what is it?

Think about the start of the golf ball's motion on my picture:



Oh, that's handy, it's a right-angled triangle! So we can find uy using Pythagoras' Theorem.

[There's another way to find the speed of the golf ball. If we look at the original graph we see that it takes 2.5 seconds for the golf ball to reach its minimum speed. Using the equation vy = uy - gt and recalling that at the top of the motion vy = 0 tells us that u=gt. Intriguingly this leads to a very slightly different answer, which tells me that they've rounded to 3 significant figures.]

Anyway, we now know how much time the ball spend travelling (5 seconds), its speeds in both the vertical and horizontal directions, and the acceleration.

For part (a) use x = ut for u = 19 m/s, t = 5 s, so that x = 95 metres. Which tells me that the guy was probably using a sand-wedge or maybe an 8-iron.

For part (b) we can use either s = uyt - (1/2)*gt2 for t = 2.5 seconds and uy = 24.5 (m/s), g = 9.8, or s = uy2/2g. These should both give an answer of 30.6 metres when rounded (by the way, uy2 is exactly 600).



The SUVAT equations haven't just been drawn out of a hat, and can be derived as follows (it's worth learning this so that you don't have to remember them off-by-heart). Apologies for the confusing notation, but there's a good week or so of work behind this, and I'm assuming you've done most of that, and especially that you have some understanding of differential and integral calculus:

Firstly, we'll look at the equations themselves. We notice that the velocity and displacement depend on the time. This can be written by saying, instead of just v and s, we can write v(t) and s(t) - v and s are functions of time.

Also, we remember that velocity is defined as (change in position)/ (time taken for change). When these changes are always happen this is written in terms of derivatives as v= ds/dt , which means "rate of change of position with respect to time".

Also we have the acceleration which is (change in speed)/ (time taken for change). Again, if the speed is constantly changing with time, then we write a = dv/dt = d/dt(ds/dt). That's the rate of change of "the rate of change of distance with respect to time" with respect to time. A bit of a mouthful.

Anyway, we're left with just one equation:

a = d/dt(ds/dt) which is written d2s/dt2.

Now supposing that the acceleration is a constant, say g (g points downwards)

Then:

d2s/dt2 = -g

We now integrate both sides with respect to time:



Where u is the velocity at time t = t0. This is our first equation written above*.

Now we do the same thing again, integrate up with respect to t, to find s:



A couple of extra terms have appeared! However in most problems the s0 and t0 have been chosen to be zero - that is, the problem starts at t = 0 and the ball starts moving from a point that is called the zero point.

So really all the equations say the same thing, that acceleration is a constant.

The final equation, v2 = u2 + 2as, comes from first squaring both sides of the equation v=u+at:

v2 = (u+at)2
    = u2 + 2atu + a2t2
    = u2 + 2a (ut + (1/2)at2)
    = u2 + 2as

* What, in that picture, does "dt' " mean, and why have I talked about t' too in the second integral? Umm...

In high school the first integrals you learn are called "indefinite", which means that there are no limits to the integral and once you've done the integral you are left with a function. Then you learn about definite integrals, with numbers as limits, and you get a number back. What I have written is the sort of crossover between the two - an indefinite integral, with a variable limit, so you get a function back with limits.

The t' is now known as a "dummy variable". That means that we don't care about it much, only the limits of the integral, which are t (varies with time, obviously) and t0 a constant. Then why have I used t'? It's just to distinguish it from the real variable, which is t.

Put another way... You don't need to know about that. I was just showing off.
« Last Edit: March 15, 2011, 04:02:00 AM by BFM_three60 »
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Offline TUR80

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Re: physics/maths question
« Reply #4 on: March 14, 2011, 08:37:14 PM »
nah, it has to do with the constant acceleration formulas,
since gravity is involved, speed will never be the same (unless a constant force is added),
and i think i just answered me own question xd

i need a good physics expert to tell me whether this is right or wrong
using the formula
s=ut + 1/2(a*t^2)
where s=0.544m, t=0.2secs, a=-9.8m/s^2 and u=initial velocity
the only thing im unsure of is the acceleration


i also still need help with the 2nd question



i love those formulars

do you know
v^2=u^2+2as
or
v=u+at

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Offline BFM_Xtr3me

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Re: physics/maths question
« Reply #5 on: March 14, 2011, 11:49:44 PM »
TYVM THREE60!!!!
i figured out the 1st one as i was typing the my last response and you posted the same method so i know im right,
ive understood most of the 2nd part so ill just re-read it till i get it,
i guess i know who to turn to for these types of questions now xd

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Offline BFM_Fénix

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Re: physics/maths question
« Reply #6 on: March 15, 2011, 09:20:12 AM »
You should check the tutors thread...
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Offline jim360

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Re: physics/maths question
« Reply #7 on: March 15, 2011, 09:47:39 AM »
more people post in the Tutors thread saying they are tutors than have ever asked for help. I guess people don't often come here on a long-term basis for tutoring. Pity, because I'd do it free.
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Offline BFM_Fénix

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Re: physics/maths question
« Reply #8 on: March 16, 2011, 06:20:15 AM »
Indeed a pity.

Even though this is a game-dedicated forum, a lot of people in here are prepared in many subjects... Also, the Tutors thread is outdated...

BTW xtreme, did you understand 360's explanation (2nd part)?
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Offline BFM_Xtr3me

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Re: physics/maths question
« Reply #9 on: March 16, 2011, 11:13:56 PM »
yes, i finally got it,
after i read it once, my lecturers covered the exact same thing the next day so i came back on and read it again and it made sense :)
























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