#1

So I did a question for Mechanics that my maths teacher couldnt do, as well as the rest of the class. It involves the basic laws of motion. Ill give you no more hints. See if you can get it.

(retardation is deccelerating)

I have the answer, if anyone wants to check what they got. Have fun.

I did

Seriously.

An underground train covers 576 m from rest to rest in 60s. At first it has a constant acceleration of 0.5 ms^-2, then moves with a constant speed and finally has a constant retardation of 1 ms^-2. Find the time taken for each stage of the journey

(retardation is deccelerating)

I have the answer, if anyone wants to check what they got. Have fun.

I did

Seriously.

#2

I would but I just had my physics final today and i'm looking forward to not having to do any more physics until September 2008

#3

I don't know, but it has an average speed of 28.8 m/s (yes I know I phail).

#4

^When you say "I know I phail" do you mean you fail because you couldn't do anything else or you fail because that answer is completely wrong?

Looks like a speed-time graph question. Can't be bothered to figure out whether it can be done another way.

Looks like a speed-time graph question. Can't be bothered to figure out whether it can be done another way.

#5

^When you say "I know I phail" do you mean you phail because you couldn't do anything else or you fail because that answer is completely wrong?

Looks like a speed-time graph question. Can't be bothered to figure out whether it can be done another way.

Velocity-time graphs do come into it, I suppose. I didnt use them, though. It just requires you to think logically.

I don't know, but it has an average speed of 28.8 m/s (yes I know I phail).

Actually, it doesnt go past 20 ms^-1 at any point.

#6

ugh gimme the a time and it'll be easier i dont wanna break out my physics notes

#7

Velocity-time graphs do come into it, I suppose. I didnt use them, though. It just requires you to think logically.

Well the amount of time for the deceleration has to be half the time of the acceleration. That's as much as I can be bothered to work out.

#8

Well the amount of time for the deceleration has to be half the time of the acceleration. That's as much as I can be bothered to work out.

How

*do*you do it?

#9

I got the first stage of the journey as 38.4 seconds and the second stage as 19.2 seconds, but that leaves .4 seconds missing from the answer. Ive used most of the ways I can think of.

Am I right in saying that the maximum speed reached is 19.2 ms^-1?

And going by what Meths said, if the deceleration is half the time of the acceleration, then its in a ratio of 2:1, ao you'd divide it by 3 to get t, then multiply by 2 to get the time for acceleration, which gets 40 and 20, nearly the same as what I got.

.

Am I right in saying that the maximum speed reached is 19.2 ms^-1?

And going by what Meths said, if the deceleration is half the time of the acceleration, then its in a ratio of 2:1, ao you'd divide it by 3 to get t, then multiply by 2 to get the time for acceleration, which gets 40 and 20, nearly the same as what I got.

.

*Last edited by SmarterChild at Jun 19, 2007,*

#10

The awnser is Pi^4

#11

I got the first stage of the journey as 38.4 seconds and the second stage as 19.2 seconds, but that leaves .4 seconds missing from the answer. Ive used most of the ways I can think of.

Am I right in saying that the maximum speed reached is 19.2 ms^-1?

And going by what Meths said, if the deceleration is half the time of the acceleration, then its in a ratio of 2:1, ao you'd divide it by 3 to get t, then multiply by 2 to get the time for acceleration, which gets 40 and 20, nearly the same as what I got.

Nowhere near. The answer is

*far*more involved than that. It took up nearly a full A4 page to work it out...

#12

What would it matter anyway?, trains or always too late or too early.

#13

^

I think I could do it but I'd need to use the equations of motion and other things I can't do in my head just looking at a screen.

Nowhere near. The answer isfarmore involved than that. It took up nearly a full A4 page to work it out...

I think I could do it but I'd need to use the equations of motion and other things I can't do in my head just looking at a screen.

#14

Oh, bugger I didnt read the question properly. Missed out the bit with the constant speed!

#15

CHUCK NORRIS!

But only cos i cant be arsed thinking about physics, having just done an A-level exam in it.

But only cos i cant be arsed thinking about physics, having just done an A-level exam in it.

#16

You didn't give enough information......

Either that or I cant be bothered to actually get a piece or paper and a pencil out....

EDIT: You retard, I know why everyone couldn't get it. You didn't give a distance for the periods of time for deceleration, constant speed and acceleration. Unless you're just assuming random stuff it IS an impossible question.

Either that or I cant be bothered to actually get a piece or paper and a pencil out....

EDIT: You retard, I know why everyone couldn't get it. You didn't give a distance for the periods of time for deceleration, constant speed and acceleration. Unless you're just assuming random stuff it IS an impossible question.

*Last edited by XMetallica73X at Jun 19, 2007,*

#17

So I did a question for Mechanics that my maths teacher couldnt do, as well as the rest of the class. It involves the basic laws of motion. Ill give you no more hints. See if you can get it.

(retardation is deccelerating)

I have the answer, if anyone wants to check what they got. Have fun.

I did

Seriously.

The question as you give it right now can't be solved, since we don't know at what time or distance the train's velocity goes constant, and then decelerates.

Edit: Or what Xmetallica73X said

#18

Yeah, I just tried solving it as far as I could and your missing the sperate distances, sorry.

We could try a Calculus equation if you like.......

Find the Area under the curve of Y=x^2-2x+3 within the interval [-2,2]

I have the answer when you're ready....

We could try a Calculus equation if you like.......

Find the Area under the curve of Y=x^2-2x+3 within the interval [-2,2]

I have the answer when you're ready....

*Last edited by XMetallica73X at Jun 19, 2007,*

#19

Damm you and your unsolvable Maths questions. I can't see a way, depending on the time taken travelling at the constant speed, there could be a whole range of answers. And I can't see a way of solving it through calculus, cos its three seperate gradients and not a smooth curve and I dont know how you'd find an equation for such a line.

I wasted some of my birthday for this question .

I wasted some of my birthday for this question .

*Last edited by SmarterChild at Jun 19, 2007,*

#20

Damm you and your unsolvable Maths questions. I can't see a way, depending on the time taken travelling at the constant speed, there could be a whole range of answers!

I wasted some of my birthday for this question .

Well happy birthday, it's unsolvable anyways.

#21

Yeah, I just tried solving it as far as I could and your missing the sperate distances, sorry.

We could try a Calculus equation if you like.......

Find the Area under the curve of Y=x^2-2x+3 within the interval [-2,2]

I have the answer when you're ready....

FUNNY S SHAPE 1/3x^3 - x^2 + 3x dx

[1/3x^3 - x^2 + 3x]-2, 2

1/3 x -2^2 - -2^2 - 6 = 1/3(4) - 4 - 6 = -10 + 1 + 1/3 = -8 + 2/3

1/3 x 2^2 - 2^2 + 6 = 1/3(4) - 4 + 6 = 3 + 1/3

-8 + 2/3 - (3 + 1/3)

= - 12 units^2

Would have been easier with paper, so I could well be wrong - but I think that's right.

EDIT: Wait, that can't be right.

You can't have a negative area.

3 + 1/3 - (-8 + 2/3)

= 11 + 1/3 - 2/3

= 10 + 2/3 units squared.

#22

I will, but only if you give a good reason as to why d/dx(arccot(x)) is always negative.

*Last edited by porcupineoris at Jun 19, 2007,*

#23

12 Todays answer is 12

If you deem that incorrect then you sir have asked the wrong question.

If you deem that incorrect then you sir have asked the wrong question.

#24

Part of the answer is 12. Remember, you have to find the rest of the times, as well.

+!p3: proof it can be solved:

+!p3: proof it can be solved:

```
First of all, let [T1]=x [T2]=y and [T3]=z
[T1,2,3]Being the time taken to accelerate, etc respectively.
From the law of motions, we can deduce:
[u1]=0
[v1]=x/2
[u2]=x/2
[v2]=x/2
[u3]=x/2
[v3]=0
Therefore, we can use tha law of motions to again deduce that:
s=(0.5([u1]+[v1])[T1]) + (0.5([u2]+[v2])[T2]) + (0.5([u3]+[v3])[T3])
simplifying, and inserting the appropriate values;
576= x²/4 + xy/2 + xz/4
2304= x² + 2xy + xz -----------[1]
z=x/2
therefore:
2304= x² + 2xy + x²/2
60=x+y+z
60= 3x/2 +y
x= 40 - 2y/3 ------------------[2]
Substitute [2] in [1];
2304= (40-2y/3)² + 2y(40-2y/3) + (20-2y/3)(40-2y/3)
After simplification:
2304= 2400 - 6y²/9
6y²=864
y²=144
y=12
Insert y in [2]
x=40-8
x=32
z=32/2
z=16
therefore:
[T1]=32s
[T2]=12s
[T3]=16s
```

*Last edited by umop-3p!sdn at Jun 20, 2007,*

#25

This a SUVAT question? God I hate them.

#26

Yeah, except its a much harder question than youd get on any exams, as it doesnt actually define all the values.

#27

Yeah, except its a much harder question than youd get on any exams, as it doesnt actually define all the values.

I guessed it would be some simultaneous equations, too many unknowns.

Cool ownage of the guys who said it was impossible though.

#28

well, in a roundabout way, it does.Yeah, except its a much harder question than youd get on any exams, as it doesnt actually define all the values.

if you go to VELOCITY as the basis for this:

you know that the

*slope*of the 3rd phase is

*twice*that of the first.

so the

*time*for the 3rd phase will be 1/2 of the first.

the time of the 2nd phase is 60sec - the time for the other 2 phases.

the area under the graph is the distance.

the area under of the first and last phases (distance) is =

**1/2**the

*maximum*velocity

**x**their respective times.

the area under the middle section is the (steady state) velocity (which is equal to the

*maximum*velocity of the other 2 phases)

**x**it's time.

It's been far too long since I used calculus, to remember how to properly substitute and come up with an appropriate expression for this, but my guess is our boy-geniuses (JamieB and XMetallica73X) probably CAN.

I simply looked at the condition where t2=0 (the

*greatest*distance possibly traveled with the total time and accelerations specified) and found it to be 600m.

Since 600m > 576m , there IS a solution.