Kinematics One Dimensional help

[toesockshoe]

Homework Statement

A train starts from a station with a constant acceleration of at = 0.40 m/s2. A passenger arrives at the track time t = 6.0s after the end of the train left the very same point. What is the slowest constant speed at which she can run and catch the train. Sketch curves for the motion of the pasenger and the train as functions of time

Homework Equations

x=x₀+v₀t+1/2at²[/B]

The Attempt at a Solution

let origin be when the person reaches the tracks (at t=6).
let h = the distance the train has already traveled in the first 6 seconds)

PERSON
x=v_{op(initial of person)}t

TRAIN
x=h+v_{ot(initial of train}+a_tt²/2

let position of both objects equal each other

v_opt=h+v_opt+a_tt²/2

what do i do now? i am stuck. i have to find v_op, but i can't seem to find out how to find t... i know the vlaues of h,v_ot, and a_t.

 

[BvU]

You want to also let the times be equal. That's the seond equation.
Make a graph of position vs time for both, easier to see what you are doing then.

 

[toesockshoe]

You want to also let the times be equal. That's the seond equation.
Make a graph of position vs time for both, easier to see what you are doing then.

i tried doing that. i did make a position vs time graph, but i didnt really know how to find a way to write equations for time without introducing a bunch of new unknown variables (like velocity of train AFTER the time driven). can you give me a hint?

 

[FallenLeibniz]

I don't believe what you have written for the equivalence between the train and the person's position is correct. Did you want to put in the initial velocity of the train on the right side first?

 

[toesockshoe]

I don't believe what you have written for the equivalence between the train and the person's position is correct. Did you want to put in the initial velocity of the train on the right side first?

oh sorry. i copied it wrong. i did it right on my paper. it should be v(ot) on the right side. it won't let me change it now.

 

[FallenLeibniz]

You have the right idea with the setting of the equivalences. You'll want to use that to get the time that the person need to run before he catches the train at the constant velocity. Once you get that, you can solve for how far he must run via how far the train has gone from the original point. Then you can get the original velocity from that. Does that help?

 

[PeroK]

i tried doing that. i did make a position vs time graph, but i didnt really know how to find a way to write equations for time without introducing a bunch of new unknown variables (like velocity of train AFTER the time driven). can you give me a hint?

It looks simpler to me to start at t=0, so that the person starts running at t = 6. This removes h and the initial velocity of the train from your equations.

 

[FallenLeibniz]

Rearrange the equivalence that you set so that you get a quadratic equation.

 

[SteamKing]

i tried doing that. i did make a position vs time graph, but i didnt really know how to find a way to write equations for time without introducing a bunch of new unknown variables (like velocity of train AFTER the time driven). can you give me a hint?

Why are you complicating this problem?

Instead of introducing a new variable for everything, try to make some variables do double duty.

For instance, there's no need to have multiple time variables. One time variable should be sufficient for this problem.

Assume t = 0 when the train starts to move. At t = 6 sec. is when the passenger shows up at the station and starts chasing the train.

Using t, you should be able to use the equations of kinematics to calculate the distance the train travels from the station. The same t, offset by 6 secs., can be used to figure out the distance the late passenger runs trying to catch the train, given a constant velocity v_p. These two curves can then be easily plotted.

 

[SammyS]

Homework Statement

A train starts from a station with a constant acceleration of at = 0.40 m/s2. A passenger arrives at the track time t = 6.0s after the end of the train left the very same point. What is the slowest constant speed at which she can run and catch the train. Sketch curves for the motion of the pasenger and the train as functions of time

Homework Equations

x=x₀+v₀t+1/2at²[/B]

The Attempt at a Solution

let origin be when the person reaches the tracks (at t=6).
let h = the distance the train has already traveled in the first 6 seconds)

PERSON
x=v_{op(initial of person)}t

TRAIN
x=h+v_{ot(initial of train [?])} [ t? ] +a_tt²/2

let position of both objects equal each other

v_opt=h+v_opt+a_tt²/2

It looks like you have some typos in the above.

That last equation should be:
v_opt=h+v_ott+a_tt²/2

what do i do now? i am stuck. i have to find v_op, but i can't seem to find out how to find t... i know the values of h,v_ot, and a_t.

If you did know v_op, how would you solve for t ?

By the way, I think your choice for time, t, is fine.

 

[FallenLeibniz]

You only need to account for the time when the train is originally moving to get the h variable (the distance it is ahead at the time the person gets there). After that, all you need to worry about is the time for which the person is running. Isn't it easier to just break the problem up like that?

 

[BvU]

v_opt=h+v_opt+a_tt²/2

You mean v_opt=h+v_ott+a_tt²/2 , right ?

Sorry, I confused you earlier on.
You made the drawing, so you can see how to wiggle the straight line for the passenger to let him/her catch the train. The minimum v₀ is when the line is a tangent, right ? I.e. when there is 1 single solution for your equation $$
v_p t = h + v_{0,{\rm train}} t + {1\over 2} a t^2
$$ (which is correct). It is a quadratic equation in t. When does that have a single solution ?

 

[toesockshoe]

It looks simpler to me to start at t=0, so that the person starts running at t = 6. This removes h and the initial velocity of the train from your equations.

i already know h. h would just be 6 times 0.4 in this case.

 

[toesockshoe]

yeah i know i made some mistakes in the notations. t stood for train and time and yeah its confusing. i changed in on my paper. the equation is now
v_opt = h+v_oct+1/2at^2 . v_oc is the initial velocity of the train at t=6.none of you guys are giving me a clear explanation how to find t. i have 2 unkowns in the above equations (v_op AND t). i plotted the xvt graph. I've attached what i drew.

 

[BvU]

Good. Lousy picture, I have to strain my neck to see it. Never mind.
v_person is the slope of the straight line. It is easy to see that a lower value of v_person still let's him/her catch the train, right ?
Up to you to find the lowest slope that still allows him/her to catch the train. Now, in what situation does that happen ?

none of you guys are giving me a clear explanation how to find t

Maybe. But in the mean time you have managed to let a whole lot of extremely knowledgeable experts go out of their way to help you further. Not bad at all !

 

[SammyS]

yeah i know i made some mistakes in the notations. t stood for train and time and yeah its confusing. i changed in on my paper. the equation is now
v_opt = h+v_ott+1/2at^2 . v_oc is the initial velocity of the train at t=6.none of you guys are giving me a clear explanation how to find t. i have 2 unknowns in the above equations (v_op AND t). i plotted the xvt graph. I've attached what i drew.

Solve for t using the quadratic formula. Yes, you will have v_op under the radical. Depending upon the value of v_op, you may or may not get a real solution for t. (This is a clue to getting a solution!)

By the Way: The graphical method others suggest is a more meaningful approach from a physical point of view.

Added in Edit: " for t "

 

[toesockshoe]

Solve for t using the quadratic formula. Yes, you will have v_op under the radical. Depending upon the value of v_op, you may or may not get a real solution. (This is a clue to getting a solution!)

but we don't know the value of Vop. would we able to find t?

 

[BvU]

Yes

 

[FallenLeibniz]

h=.4*6 is dimensionally incorrect. h is supposed to have dimensions of length. Why would you multiply the acceleration by the time to get a length? You need to integrate.

I pose this question now...
The velocity of the person should be equal to the train at the time that he catches it. True or false (forgive me, but I'm a bit rusty which is why I'm asking)?

If it is true, you can actually put the velocity of the person at the time that he catches the train in terms of the time at which he catches the train and the acceleration. Once you have that, then you can plug that into the quadratic you get by setting the positions equal to each other. @toesockshoe : Is that a helpful tidbit?

 

[toesockshoe]

h=.4*6 is dimensionally incorrect. h is supposed to have dimensions of length. Why would you multiply the acceleration by the time to get a length? You need to integrate.

I pose this question now...
The velocity of the person should be equal to the train at the time that he catches it. True or false (forgive me, but I'm a bit rusty which is why I'm asking)?

If it is true, you can actually put the velocity of the person at the time that he catches the train in terms of the time at which he catches the train and the acceleration. Once you have that, then you can plug that into the quadratic you get by setting the positions equal to each other. @toesockshoe : Is that a helpful tidbit?

sorry i was typing that response while i was out. yeah h would be 1/2(0.4)*(6^2)

and false, the velocity of the person and train are NOT going to be equal when the person catches up.

i just came home and will solve the quadratic formula.

 

[PeroK]

and false, the velocity of the person and train are NOT going to be equal when the person catches up.

You should probably finish the problem the way you started it. But, actually, the person and train will be going at the same speed when she catches up. That gives, in fact, a simpler way to solve the problem, as it avoids the quadratic equation.

 

[toesockshoe]

h=.4*6 is dimensionally incorrect. h is supposed to have dimensions of length. Why would you multiply the acceleration by the time to get a length? You need to integrate.

I pose this question now...
The velocity of the person should be equal to the train at the time that he catches it. True or false (forgive me, but I'm a bit rusty which is why I'm asking)?

If it is true, you can actually put the velocity of the person at the time that he catches the train in terms of the time at which he catches the train and the acceleration. Once you have that, then you can plug that into the quadratic you get by setting the positions equal to each other. @toesockshoe : Is that a helpful tidbit?

k yhanks

 

[SammyS]

...
and false, the velocity of the person and train are NOT going to be equal when the person catches up.

So that the person will be going faster than the train when he/she catches up.
... but doesn't that mean that the person could have caught the train if he/she had been a little slower?

 

[toesockshoe]

j

 

[toesockshoe]

So that the person will be going faster than the train when he/she catches up.
... but doesn't that mean that the person could have caught the train if he/she had been a little slower?

wait what? the person and train are at the same speed when the person catches up? really?

 

[toesockshoe]

So that the person will be going faster than the train when he/she catches up.
... but doesn't that mean that the person could have caught the train if he/she had been a little slower?

if the person isn't going faster than the train how can the person make up for the 6 second headstart the train had? doesn't this mean that the persons speed HAS to be greater than the trains?

 

[FallenLeibniz]

As for the speed thing, if you take the time derivative of both sides after setting the two position functions equal, you can see this. Keep in mind that the velocities only match that for the instant at which his position function catches up with that of the train. A good qualitative argument is the following:

The person must start out with a velocity that is faster than that of the train at the time he starts running. If he is slower, he never catches up to the train and he can never run with the train since his speed must be constant from the start. However, since the train is accelerating, the train's velocity will keep increasing toward the person's. Thus, if he starts running at the "right" velocity that is more than the train, the train's velocity curve will "catch up" with his then pass it. That "catch up" moment just so happens to be when the positions are equal.

You are actually right that you can just use the kinematics equations for h (I was thinking of integrating because of the acceleration, but your way is simpler). However, the equations you need is $h=\frac{1}{2}at^2$ with t here being the six seconds.

 

[toesockshoe]

So that the person will be going faster than the train when he/she catches up.
... but doesn't that mean that the person could have caught the train if he/she had been a little slower?

alright.

so let me solve the problem.

the position functions set equal to each other:

h+v_otraint+(1/2)at²=v_opersont

since the velocity of the train and person are the same when they catch up:
v_operson=v_ftrain
v_ftrain=v_otrain+at
so v_operson=v_otrain+at

plug the above equation into the formula 5 lines above...

h+v_otraint+(1/2)at²=(v_otrain+at)t
h+v_otraint+(1/2)at²=v_otraint+at²
v_otraint cancels out
so we get
h+(1/2)at²=at²
plug in 7.2 for h, 0.4 for a, and you get t=6 seconds. this isn't right... the answer says 4.8. can you tell me where i am going wrong?

 

[FallenLeibniz]

First off, let me start by saying that I am a complete and total moron and I apologize for my idiocy..

My solution forgot to take into account the fact that when the person starts running, the train has an initial velocity that is non zero.

 

[FallenLeibniz]

Found out what you're doing. You're plugging in the time value as an answer for velocity. You need to plug it into the velocity for the person. You get 4.8 m/s then.

 

[toesockshoe]

ugh, I am a moron too. btw, can you guys tell me what difficulty you would think this problem is? i would never have figured out the velocities equal each other if no one pointed it out... ugh. i feel like this is too hard a course for me.

 

[FallenLeibniz]

It's not exactly a hard one but not a piece of cake either. Don't worry. Textbooks are filled with a good amount of ones like these. The best thing to do would be to practice and/or study with a friend. You can do it. Just got to keep on truckin.

It may be helpful to keep units in your numbers just for awhile. The time goof could have been avoided if it were clear that you were talking about seconds and your answer was talking about meters/second. I do it too, and got confused too because I wasn't watching the units.

Also, I would suggest breaking the problems up into pieces and solving each piece first rather than getting up all of your substitutions into one big thing. It may make it easier to find any mistakes if you lay it out like this as well as easier to go through your work.

 

[PeroK]

ugh, I am a moron too. btw, can you guys tell me what difficulty you would think this problem is? i would never have figured out the velocities equal each other if no one pointed it out... ugh. i feel like this is too hard a course for me.

I think you chose a complicated way to do it. If you begin at t = 0 when the train starts to move, you get:

$x_t = \frac{1}{2}at^2$

$x_p = v(t-6)$

This is much simpler than your equations.

With this approach, you are looking for v where you can solving the quadratic equation in t (by setting $x_t = x_p$). And, to get the minimum v, you need the quadratic to have exactly one solution:

No solutions => v is too low and she never catches the train
Two solutions => v is too high and she catches up with the train, overtakes it, but it keeps speeding up and then catches her (so there are two times where the distances are equal)
One solution => v is just right and she just catches the train (just as the train reaches her speed)

There are, of course, other ways to solve the problem. Thinking about the equal velocities is a different solution. I wouldn't worry about that for now. I would concentrate on solving the quadratic equation implied above.

The key question is: when does a quadratic equation have exactly one solution?