help lark pass physics

[Kestrel]

Suggested in the bad day thread, moving all my physics woes here.

The next test is in a few days and I have no idea how to do even the basic concepts. This has to do with speed, velocity, acceleration, etc, graphs based on them, and word problems. I'm having trouble even with what was taught on day one. I hate.

Anyway.

Examples of the simple problems I should know how to do but don't:

-- Lance Armstrong decides to go for a bike ride. He starts at a speed of 10 m/s, but then decides to go downhill. Downhill he speeds up at a constant rate. His overall descent is 150m and takes him 10s.

------- How fast is he going at the end?

------- By what amount did his velocity increase every second?

-- You hike 2/3 up a mountain at a speed of 3mi/hr and run the final third at 6mi/hr. Average speed?

-- a horse moves away from its trainer in a straight line, 116m away in 14s. It turns and gallops halfway back in 4.8s. Calculate average speed, and average velocity, away is positive.

Some more complex things:

-- A falling stone takes .28s to travel past a window 2.2 m tall. From what height above the top of the window did it fall?

I'll put more up later, my hand hurts but yeah. I'm so confused.

[+Slowswift]

Suggested in the bad day thread, moving all my physics woes here.

The next test is in a few days and I have no idea how to do even the basic concepts. This has to do with speed, velocity, acceleration, etc, graphs based on them, and word problems. I'm having trouble even with what was taught on day one. I hate.

Anyway.

Examples of the simple problems I should know how to do but don't:

-- Lance Armstrong decides to go for a bike ride. He starts at a speed of 10 m/s, but then decides to go downhill. Downhill he speeds up at a constant rate. His overall descent is 150m and takes him 10s.

------- How fast is he going at the end?

------- By what amount did his velocity increase every second?

-- You hike 2/3 up a mountain at a speed of 3mi/hr and run the final third at 6mi/hr. Average speed?

-- a horse moves away from its trainer in a straight line, 116m away in 14s. It turns and gallops halfway back in 4.8s. Calculate average speed, and average velocity, away is positive.

Some more complex things:

-- A falling stone takes .28s to travel past a window 2.2 m tall. From what height above the top of the window did it fall?

I'll put more up later, my hand hurts but yeah. I'm so confused.

 

I'm afraid I won't be any help, as just reading that makes my head hurt, too.

 

However, have you heard of Wolfram Alpha? That might help, if you can't find anything here.

 

Good luck!

Edited September 12, 2015 by Slowswift

[ParadoxicalZen]

In the 3 month gap between finishing secondary school and starting college, my B grade GCSE went to non existent. Sometimes i struggle to remember that 2+2 = 4...or 5, i forget x) Seriously, if had time to go over all the stuff I learnt at school again, I'd happily help but this looks like Rocket Science. In my day we just had simple sums >.>

[TCshard]

-- Lance Armstrong decides to go for a bike ride. He starts at a speed of 10 m/s, but then decides to go downhill. Downhill he speeds up at a constant rate. His overall descent is 150m and takes him 10s.
------- How fast is he going at the end? 15m/s

150/10=15
------- By what amount did his velocity increase every second? .5m/s he sped up 5m/s in 10s, thats half a meter per second.

 

-- You hike 2/3 up a mountain at a speed of 3mi/hr and run the final third at 6mi/hr. Average speed?

You first divided the hike into thirds

You hike 1/3 at 3 mph, you hike 1/3 at 3 mph, you hike 1/3 at 5 mph.

3+3+6=12

12/3=4

This is simply knowing what to divided

 

-- a horse moves away from its trainer in a straight line, 116m away in 14s. It turns and gallops halfway back in 4.8s. Calculate average speed, and average velocity, away is positive.

116/2=58

116+58=174

first average speed is 8.29(rounded)

second average speed is 12.08(rounded)

You once again divide into three parts, 1/3 at 8.29, 1/3 at 8.29, 1/3 at 12.08

2(8.29)+12.08=28.66

28.66/3=9.55(rounded)

9.55s to run 58m

6.07 m/s

I think I may be wrong, I rounded a lot.

 

Some more complex things:
-- A falling stone takes .28s to travel past a window 2.2 m tall. From what height above the top of the window did it fall?

Edited September 12, 2015 by TCshard

[Haelbarde]

I started to go through the first question, but megaly over complicated it. I've put what I had in spoilers, in case it becomes relevant. Otherwise, just ignore the spoilers.

This is where I wish the editor had LaTex/Mathprint support. 

 

I'm presuming this is year 11 / stage 1 / Senior Physics (what ever it's called?).

I'd start looking at relevant equations:

 

s(t) = vo * t + 0.5 * a * t^2

v(t)= vo +at

 

For that first part, what we want is v(10) {i.e. his speed/velocity after 10 seconds).

Now, from the question we have vo (the initial velocity) as being 10 m/s, and we already know we have t (=10 s). So from that second equation, all we need to solve the problem is a (acceleration).

 

How do we find that?

Well, my first thought would be F=ma, but we have neither F, nor m. But notice we have a distance given in the question. Do we have equations that involve distance? 

 

Yes we do. That first one. 

So what exactly is that distance? It says 'overall descent'. I would take that to refer to the vertical displacement. So we have s_v(t) {s subscript v, for vertical displacement) = 150 m. Considering that first equation in the vertical axis, we have:

 

s_v(t)= vo_v * t + 0.5 * a_v * t^2

 

Putting in the values we know, we have:

-150 = vo_v *10 +0.5*a_v*(10)^2 {I've made the displacement negative because he's going down hill)

What's vo_v? It's the inital vertical velocity. Let's look back at the question. Initially, he was cycling on level/flat ground, so until he started going down the hill, his vertical velocity was 0.

 

-150 = 0.5*100*a_v

-150=50*a_v

a_v=-3 m/(s^2)

 

 

And I've run out of time for them moment. Do you know how many marks that first question would be worth?

 

For that second one, you spent 2/3rds of the hike walking at 3 mi/hr, and 1/3rd at 6mi/hr. Think that'll just be (2/3)*3 + (1/3)*6 = 4 mi/hr. You can think about it like this: Say the hike is 3 hours. 2 hours were spent walking at 3 mi/hr, so over that 2 hours, you'd have travelled 6 miles. Then for the last hour, you traveled 6 mi/hr, so traveled a distance of 6 miles. Thus you traveled 12 miles in total, and it took you 3 hours to do so. 12/3 = 4mi/hr. Make sense?

 

If I have time later, I'll jump back on to help answer/explain things. Feel free to PM me too.

[ccstat]

What type of responses would be helpful to you? Is it the concepts you have trouble with? Converting the problems to equations? Doing the math after you have the equations? All of the above? I'm happy to help where I can, but just giving you the set of equations you need and the answer may not be what you want. 

Also, are you using arithmetic or basic calculus for these? It is possible both ways, but sometimes the calculus is actually easier if you know how to use it. For now I'll assume arithmetic only.

 

I don't have a lot of time now, but I'll be on a bit later to explain a problem or two.

 

For that second one, you spent 2/3rds of the hike walking at 3 mi/hr, and 1/3rd at 6mi/hr. Think that'll just be (2/3)*3 + (1/3)*6 = 4 mi/hr. You can think about it like this: Say the hike is 3 hours. 2 hours were spent walking at 3 mi/hr, so over that 2 hours, you'd have travelled 6 miles. Then for the last hour, you traveled 6 mi/hr, so traveled a distance of 6 miles. Thus you traveled 12 miles in total, and it took you 3 hours to do so. 12/3 = 4mi/hr. Make sense?

 

This is not quite correct. As I read the problem, you spend 2/3rds of the distance walking at 3 mi/hr, not 2/3rds of the time, which is what Haelbarde's solution uses. Here is the approach I would use: 

The total distance doesn't matter. You can pick any number and you'll get the same answer, but I'll use 18 miles because it makes the math work out nicer. 

 

The first leg is:

2/3*18mi = 12 miles. 

12 miles /3miles/hr = 4 hours

 

The second leg is:

1/3*18mi = 6 miles.

6 miles /6miles/hr = 1 hour

 

Total time is 4+1=5hours

Total distance is 18 miles

Average speed for total trip is 18 miles/5 hours = 3.6 mi/hr

[Kestrel]

Wow, thanks guys! Eep, answers aren't too much help to me, because the teacher actually puts up the answer key.

I'm not sure if its calc based but I'm assuming not, as I'm in precal this year too. Its mostly the concepts that are tripping me up, when to use what equation. After that, its simple algebra that I can accomplish very easily.

My teacher explains it poorly and I just end up confused.

[ccstat]

-- a horse moves away from its trainer in a straight line, 116m away in 14s. It turns and gallops halfway back in 4.8s. Calculate average speed, and average velocity, away is positive.

 

 

As you probably know, both speed and velocity are measures of how far you go in how much time. This question is asking if you know the difference in what they mean. Average speed doesn't care where you end up, just how fast you go the whole time you are traveling to get there. Average velocity, on the other hand, doesn't care how you travel, just where you end up and how long it takes. 

 

Let's do velocity first: at the end, the horse is half of 116m away from where it started, or 58m. The total time elapsed is 18.8s. So average velocity to get there is 58m/18.8s=3.085 m/s (round to 3.1 using two significant figures).

Now speed: at the end, the horse has traveled 116m, then another 58m, for a total of 174m. Once again, the total time elapsed is 18.8s. So average speed is 174m/18.8s=9.255 m/s (round to 9.3 m/s using two significant figures). 

 

 

If you were to calculate the velocity of step 1 (116m/14s = 8.3 m/s) and step 2 (-58m/4.8s=-12m/s) you can see why you get these kinds of numbers. Speed is the same as velocity but without caring about direction, so the negative on the -12m/s gets ignored for the speed problem, but included for the velocity problem. The average of 8.3 and -12 (weighted by how long each one happened) is going to be somewhere between them, so a number like 3.1 makes sense for velocity. On the other hand, a similarly weighted average of 8.3 and 12 will be between those two positive numbers, so 9.3 is appropriate.

[RippleGylf]

Do you understand vectors, and the difference between velocity and speed? Vectors are essential for most physics problems, I find.

[Kipper]

In case you don't, a vector is simply a measurement with a direction attached to it. Velocity corresponds to vectors. Whenever a problem deals with direction, you must use velocity.

Speed, on the other hand, does not have a direction.

So, in the horse problem, you have to do the problem two ways.

The first distance the horse goes can be represented like this:

-------------------->

Then he goes halfway back. Now, when finding average velocity, because the horse went halfway back, the problem will graphically look like this:

Path A:

-------------------->

Path B:

<-----------

Because the horse is going the opposite direction, you must subtract the Path B vector from the Path A vector, leaving you with a final displacement vector of:

------->

Which is half the distance, pointing to the right, or positive. This is very important! Always include a direction in velocity problems. At this point, all you have to do is plug the equation for average velocity in.

For speed, you simply add Path B to Path A, because speed doesn't care about direction. Then you plug average speed in and you don't need a direction.

Sorry for any errors, this was all typed up on mobile. Feel free to PM me.

Edited September 13, 2015 by Kipper

[RippleGylf]

Basically that. My physics teacher refers to the result as the "shortcut," as in, if you go in from point A to point B, and then from point B to point C, the result is the shortcut from point A to point C, forming a triangle.

[maxal]

Wow, thanks guys! Eep, answers aren't too much help to me, because the teacher actually puts up the answer key.

I'm not sure if its calc based but I'm assuming not, as I'm in precal this year too. Its mostly the concepts that are tripping me up, when to use what equation. After that, its simple algebra that I can accomplish very easily.

My teacher explains it poorly and I just end up confused.

 

When to use which equations is not always obvious. What has always helped me was to start each problem by writing what I know. I would write DATA on the top of the answering page and I would list all the problems tells me.  Once the data is properly listed and deconstructed, I would then see what I can calculate based on the applicable equations. Usually the answer is somewhere within and you work it up, it becomes more obvious.

 

Also, the more you write, the more points you do get, even if you get some of it wrong. 

 

It feels as if you know your equations, you know your algebra, but you have a hard time putting them together. It is a very common issue. Seriously, many people struggle to do this, even in higher levels. My best advice is to try to deconstruct the problem, don't simply wait for the answer to pop out. Work on it. Write it down. If it helps, also write down the equations you have been taught. It will help you see a pattern. See it as a puzzle, you have your data, you have your equations: which ones can you apply should be your number one questions. Also, the more problems you will do, the more at ease you will be. 

 

Never submit a blank page. You can always write something. You can always write the DATA. You can always calculate something, even if it may not be the answer or even if it does not look as if it is the answer. You'd be surprised at how many points you may get for being close, but not right on the answer. 

 

What I have often seen is people read the problem, not being able to see how to calculate the answer right away and being stuck there, no truly knowing what to do. This is when the step-by-step procedure I tried to explain comes into play. 

[Kestrel]

We haven't covered vectors yet, so let's leave that out as I don't want to get more confused (I do understand their concept because freshmen engineering turned into a physics class) but I don't remember fully how to do them.

So draw pictures, write down knowns, see what you can find?

I'm also confused on when speed/velocity average is supposed to be weighted. We covered that, and my teacher doesn't make any sense and I get confused on when to do the weighted equation or not.

[ccstat]

I'm also confused on when speed/velocity average is supposed to be weighted. We covered that, and my teacher doesn't make any sense and I get confused on when to do the weighted equation or not.

The good news is that if you aren't sure what equation to use you can ignore the whole issue by simply finding the final distance (total for speed, net for velocity) and dividing by total time.

The reason for which equation you need is actually fairly straightforward: the average is always weighted according to the time spent at each velocity. If the problem tells you the time for each velocity, you are good to go. If it tells you the distance traveled at each velocity you need to convert that to time using the weighting equation.

Going the other way, from average to partial trip velocity, can be trickier, but the same principle applies. I'll do that window example to demonstrate that when I'm on later, if someone else doesn't get too it first

[twelfthrootoftwo]

It's often helpful to see multiple approaches, so you can choose a method that works for you. This is how I like to break down physics problems:

 

So draw pictures, write down knowns, see what you can find?

Pictures are worth a thousand words, and maybe 20 equations. Let's use your window question:

 

 

A ball is released from rest, and falls (acceleration a) a distance x₁ in a time t₁. Then it passes the top of the window; by now it's travelling at speed v. It passes the window (distance x₂) in time t₂.

 

Things you know:

x₂ (window height, 2.2 m)

t₂ (time to pass window, 0.28 s)

a (acceleration due to gravity, 9.81 ms^-2)

 

Things you need to find:

x₁ (release height above window)

 

Things that might be helpful to get from one to the other:

t₁ (time between release and arrival at window)

v (speed at top of window)

 

You'll need to use v and t₁ to find x₁, and these two equations (PM/ask for explanation or derivation):

because they relate the things you know and the things you need to find.

 

To start with, let's just consider what happens at the window, and try to fit it into the second equation. The ball arrives with speed v (v₀ in the equation), falls a distance x₂ (x in the equation), in a time t₂ (t in the equation). We have one equation and one unknown, which means we can find v:

 

 

Now we need to get from v to x₁. Before we just considered what happened at the window; now we'll just consider what happens above the window. The ball is released from rest (v₀=0) and accelerates (a), falls an unknown distance (x₁) in an unknown time (t₁), and ends up at speed v.

We use the first equation, with three knowns (v,a,v₀) and one unknown (t=t₁) to get t₁:

Now we have three knowns to go into the second equation (v₀,a,t₁) to get the final result, x₁.

 

So, for these more complex problems:

1) Draw a diagram.

2) Put the information from the question into mathematical form. (In this problem, for example, you need to use the time and distance they give you, but also identify that there's a time and distance that happen above the window too.)

3) Add anything you figure out to your diagram

4) Make a reference bank of equations you think might be related to the problem. (Here, for example, any equations of motion that involve acceleration.)

5) Look for equations with exactly one unknown. You can figure out that variable's value; now it's known.

6) Create a chain of equations that get you from what you're given to what you need to find.

7) Solve the equations.

 

(It's possible to use multiple equations with multiple unknowns, but it's more complicated and less intuitive.)

[ccstat]

-- Lance Armstrong decides to go for a bike ride. He starts at a speed of 10 m/s, but then decides to go downhill. Downhill he speeds up at a constant rate. His overall descent is 150m and takes him 10s.

------- How fast is he going at the end?

------- By what amount did his velocity increase every second?

Because the acceleration is constant, the average speed is going to be the average of his starting and final speeds. (I suspect that this test will not include any non-constant accelerations, so for the purposes of studying now, you can probably assume that is true every time you deal with acceleration.)

 

Average speed = total distance / total time = 150m /10s = 15m/s

Starting speed = 10m/s (given)

Average speed = (Final speed + Starting speed)/2, so rearrange to get: Final speed = 2*15-10 = 20m/s

So, his speed increased by 10m/s over 10s, which means 1 m/s every second since it is constant.

 

-- A falling stone takes .28s to travel past a window 2.2 m tall. From what height above the top of the window did it fall?

There are a few routes to take to the solution. I'll show the one that seems intuitive to me.

Average speed across window= 2.2m/.28s = 7.86 m/s

The stone starts a bit slower and ends a bit faster than that, but it will be going that average speed when it is halfway down the window (1.1m)

Acceleration due to gravity is 9.8 m/s every second, so it took 7.86/9.8 = 0.80 seconds to get going that fast. 

The stone started at rest (0 m/s), so the average velocity between where it started and the middle of the window is (0+7.86)/2 = 3.93 m/s

Average velocity = total distance /total time, so substitute in our numbers so far: 3.93 m/s = <distance to middle of window>/0.80s

That means the distance it fell to the middle of the window is 3.93m/s*0.80s=3.14 m.

The distance it fell to the top of the window is 1.1m less than that, or 2.04 m (round to 2.0)

 

Is this making sense? I can add more words to explain things that don't, or try another approach.

[RippleGylf]

We haven't covered vectors yet, so let's leave that out as I don't want to get more confused (I do understand their concept because freshmen engineering turned into a physics class) but I don't remember fully how to do them.

So draw pictures, write down knowns, see what you can find?

I'm also confused on when speed/velocity average is supposed to be weighted. We covered that, and my teacher doesn't make any sense and I get confused on when to do the weighted equation or not.

 

Vectors was one of the first things my class did, and when it comes to velocity problems, it makes it so much easier to understand.

[ccstat]

So how did the test go?

[Silverblade5]

Two cars are traveling on a desert road, as shown below. After 5.0 s, they are side by side at the next telephone pole. The distance between the poles is 82 m.

 

 

Two cars are traveling on a desert road, as shown below. After 5.0 s, they are side by side at the next telephone pole. The distance between the poles is 82 m.

 

(a) the displacement of car A after 5.0 s
  m
( the displacement of car B after 5.0 s
  m
© the average velocity of car A during 5.0 s
 m/s
(d) the average velocity of car B during 5.0 s
 m/s

 

 

So far, I've figured out that Db=Da+82 and that Tb-Ta=5. After that though, we hit a wall. Any ideas on what to do next?

[Seonid]

So, assuming that the poles are evenly spaced (which I think can be assumed from the problem), we have this:

 

Car A has traveled 82 m in 5.0 seconds. (It has traveled from one pole to the next one, a distance of 82 m).

 

Car B has traveled 164 m in 5.0 seconds. (It started one pole behind car A and ended up side by side).

 

This should be our displacements for each vehicle.

 

The velocities are found by dividing the displacement by the time it took to travel that far. For both cars, this is 5.0 seconds. I'll leave the actual calculation to you folks.

 

Let me know if this actually helps.

[Kestrel]

Vectors was one of the first things my class did, and when it comes to velocity problems, it makes it so much easier to understand.

 we're covering them now, actually. 

So how did the test go?

-shakey laughter followed by crying-

I had it handed to me, I did.

[ccstat]

 -shakey laughter followed by crying-

I had it handed to me, I did.

I'm sorry to hear that.

(Also, thanks for the context. Without the laughter/crying I might have interpreted that as a bit of a pun, in that the teacher physically handed the test to you, so maybe you did fine...?)

Do you have someone near you that can do a bit of tutoring? There is really a limit to the utility of text explanations over a forum interface. 

I hope the next unit goes better for you! If this was helpful at all and you need us again, we're happy to help.

[Silverblade5]

we're covering them now, actually. -shakey laughter followed by crying-

I had it handed to me, I did.

What's your skype?