Question

Can somebody help me?

Answer 1

?

Answer 2

only if you tell us what to help you with

Answer 3

3 or 4?

Answer 4

Let's start with the 3rd.

Answer 5

Okay, do you think it is fair to say it should only accelerate for a certain period of time and only decelerate for the remaining part? You don't think it should be coasting at any point in time, do you?

Answer 6

Coasting means it doesn't accelerate or decelerate.

Answer 7

Suppose the optimal solution had coasting, I think we could find an even more optimal solution by accelerating and decelerating during that coasting period.
That makes sense to me intuitively at least.

So I think up to some time \(t\) it is always accelerating and for the remaining period it is just slowing down.

Answer 8

Does that sound fair to you @Yttrium ?

Answer 9

Yes. That is actually what I am thinking.

Answer 10

Now I think we really should use energy to solve this.

Answer 11

Ooopss. We haven't studied energy yet. I think the solution required is just under kinematics..

Answer 12

Wow. Ok....

Answer 13

Well integrate constant acceleration twice to get \[
d=\frac 12 at^2+v_0t+d_0
\]We know initial velocity and distance are \(0\) and acceleration is \(20\)

Answer 14

So \[
d=10t^2
\]This is our equation for distance while accelerating.
Our equation for velocity is \[
v=20t
\]
Do these equations make sense to you?

Answer 15

@Yttrium

Answer 16

Yes. Yes.

Answer 17

Now let \(t_1\) be the amount of time that passes until we start braking.\[
d_1=10t_1^2\quad v_1=20t_1
\]

Answer 18

We'll use the distance equation again: \[
d=\frac 12 at^2+v_0t+d_0
\]However, this time our initial velocity is the velocity we got to at \(t_1\) and our initial distance is similar. Thus: \[
d=\frac{1}{2}(-100)t^2+(20t_1)t+(20t_1^2)
\]

Answer 19

Why do we substitute the deceleration value?

Answer 20

We also need a velocity equation as well: \[
v=-100t+v_0 = -100t+(20t_1)
\]

Answer 21

@Yttrium Because we are coming up with the distance equation for the period of time when we are decelerating.

Answer 22

Now there are two more puzzle pieces we need:
The train is at rest when it reaches the station so \(v=0\)
and \(d=200,000\)
When the time reaches the end of the journey.

Answer 23

Thus\[
0=20t_1-100t\implies t=\frac 15 t_1
\]

Answer 24

Okay, so remember that \(t\) here is the time that has passed since pressing the breaks, NOT the time that has passed from the very start of the journey.

Answer 25

The total time is \(t+t_1\)

Answer 26

Anyway since \(t_1=5t\) we can put it into our distance equation \[
\begin{split}
200,000&=\frac 12(−100)t^2+(20t_1)t+(20t^2_1)\\
&= -50t^2+20(5t)t+20(5t)^2\\
&=-50t^2+100t^2+500t^2\\
&=550t^2
\end{split}
\]

Answer 27

Solving for \(t\) gives us \[
t=20\sqrt{\frac {10}{11}}
\]And \[
t_1+t = 5t+t=6t = 120\sqrt{\frac{10}{11}}\approx 114s
\]Since this is off quite a bit, I think I must have made an algebraic error.

Answer 28

Wait. Let's recheck.

Answer 29

Oh I put \(20t_1^2\) instead of \(10t_1^2\) for initial distance.

Answer 30

But you've changed it, right?

Answer 31

Ohhh. I already see it.

Answer 32

Correcting: \[
\begin{split}
200,000&=\frac 12(−100)t^2+(20t_1)t+(10t^2_1)\\
&= -50t^2+20(5t)t+10(5t)^2\\
&=-50t^2+100t^2+250t^2\\
&=300t^2
\end{split}
\]

Answer 33

Solving for \(t\) gives us \[
t=20\sqrt{\frac 53}
\]And \[
t_1+t = 5t+t=6t = 120\sqrt{\frac53}\approx 155s
\]This makes more sense.

Answer 34

Yeah. Sorry for asking this. But this is really my weakness. Can you teach me how you were able to answer this? I mean, how did you apply your interpretation into equations.

Answer 35

Okay. I think I have to do it on my own. Can we just proceed to the 4th question?

Answer 36

Sorry It's a bit late.

Answer 37

Can you answer my 1st question. :) I really want to learn.

Answer 38

You want to know how I analyzed the problem?

Answer 39

Yes. If you can teach me?

Answer 40

Problem analysis just takes time to get good at.

Answer 41

There are only heuristics.

Answer 42

The heuristics are:
1) Come up with variables for the things which are unknown.
2) Come up with all the equations that you can think of, and don't stop until you have as many equations as you have unknowns.

Answer 43

And the rest will be mine, I think. :)

Answer 44

So in this case we have \(t_1\) and \(t\) for the time accelerating and decelerating.

Answer 45

And the two big equations we used where the position function (when decelerating) and the velocity function (when decelerating)

Answer 46

Every equation I showed you worked towards finding some unknown value.

We found the distance traveled while accelerating with the position function (when accelerating)

Answer 47

Some times it helps to come up with another unknown if it comes with its own equation.

Answer 48

So don't be afraid to have many variables.

Answer 49

Yes. I believe it in. It's just sometime physics really make me so stupid. >.<

Answer 50

If you do this, the only dangers left are:

algebra errors - when you miscalculate
analysis errors - when you confuse one unknown value with another value.

Answer 51

For example... if I had confused \(t\) to be the time of the whole trip rather than just the time spend decelerating, then this would be an analysis error. I would have said the time it took was 26 seconds and would be way off.

Answer 52

I actually almost made this error, but I corrected myself in time. That is why I bring it up.

Answer 53

Yeah. Maybe my most error here is the interpretation itself. Is practice a good teacher?

Answer 54

The algebra error I made was I substituted \(20t_1^2\) when it should have been \(10t_1^2\).

Answer 55

Practice is the only way to improve analysis skill.

If you become more confident and aggressive, you defeat problems easier.

Answer 56

Yeah. And I didn't see that, quickly.

Answer 57

Yeah. But my problem is I can't find problems. My sources are limited. :((

Answer 58

You don't have a physics text book? Usually they have more than enough problems.

Answer 59

Actually I don't have. :(( I'm just borrowing form the library.

Answer 60

Well you might be able to download on off the internet.

Answer 61

Can we proceed to the 4th question?

Answer 62

Let's be quick then.

Answer 63

Let's come up with a height function for both balls.

Answer 64

For ball 1\[
h_1=\frac 12 at^2+h
\]For ball 2 \[
h_1=\frac 12 at^2+v_0
\]

Answer 65

We need to find that \(v_0\)

Answer 66

It is the velocity of the ball going upward.

Answer 67

We should be able to find it in terms of \(a\) and \(h\).
do you follow?

Answer 68

Ooops. Can you explain where you get those two formulas?

Answer 69

wait, I should have put \(v_0t\)

Answer 70

For ball 1\[
h_1=\frac 12 at^2+h
\]For ball 2 \[
h_1=\frac 12 at^2+v_0t
\]

Answer 71

Okay, so ball 1 is the ball that is dropped LATER on actually.
It's initial position is \(h\), initial velocity is \(0\) and it's acceleration is just that of gravity.

Ball 2 was dropped BEFORE.
It's initial height is \(0\) since it is on the ground. It's initial velocity the velocity upward after bouncing.

Answer 72

Both come from: \[
d=\frac 12 at^2+v_0t+d_0
\]

Answer 73

This is my problem. :((

Answer 74

You don't know this formula?

Answer 75

I know but the interpretation itself makes me crazy.

Answer 76

Why does the 2nd ball has no height? Does the problem mean that is it not dropped?

Answer 77

Well it has no height because it just hit the ground and bounced up

Answer 78

When we initially start out, the second ball has already hit the ground and is about to bounce up and the first ball is going to be dropped.

Answer 79

Anyway, the final velocity of something which had been dropped at height \(h\) will be \[

v_f=\sqrt{2ha}
\]where \(a\) is gravity acceleration.
Do you want me to derive this?

Answer 80

So for our second ball, it's initial velocity will be \[
-\frac 34 \sqrt{2ha}
\]The negative is there because bouncing switched the direction of velocity, and the \(3/4\) is there because it says it is only \(75\%\) of what is used to be.

Answer 81

We can update our equations now:

For ball 1\[
h_1=\frac 12 at^2+h
\]For ball 2 \[
h_1=\frac 12 at^2-\left(\frac 34 \sqrt{2ha}\right) t
\]

Answer 82

Now we set them equal to find out when their heights are the same: \[
\frac 12at^2+h=\frac 12at^2-\left(\frac 34 \sqrt{2ha}\right) t
\]

Answer 83

So: \[
\left(\frac 34 \sqrt{2ha}\right) t+h=0
\]

Answer 84

Why do we need to find out if their heights are the same?

Answer 85

No, we know their heights will be the same when they collide.

Answer 86

We need the time at that point.

Answer 87

Okay. I got it. I thought what I think is wrong. Continue. :)

Answer 88

What will happen after this? O.o

Answer 89

Well I'm getting that: \[
t=-\frac{4}{3}\frac{h}{\sqrt{2ha}}=-\frac{2\sqrt 2}{3}\frac{\sqrt h}{a}
\]

Answer 90

Hmmm, well, we know \(a=9.8 m/s\) because it is gravity but...

Answer 91

Ooops should have put:\[
t=-\frac{4}{3}\frac{h}{\sqrt{2ha}}=-\frac{2\sqrt 2}{3}\frac{\sqrt h}{\sqrt{a}}
\]

Answer 92

The fact that we still have an \(h\) component is troubling since the question almost implies that it is independent of \(h\).

Answer 93

Now since \(a=9.8m/s^2\) then \[
\frac{1}{\sqrt{a}}
\]Will have units of \(s/\sqrt m\)

Answer 94

So my answer makes sense in terms of dimensional analysis... I wonder where I went wrong.

Answer 95

Maybe I'm not even wrong.

Answer 96

Clearly something has to get rid of the \(m\) dimension in acceleration, so it can't be just dependent on acceleration.