Kinematics Problem: Finding the Minimum Running Speed to Catch a Moving Train

[zero_infinity]

Homework Statement

A train pulls away from a station with constant acceleration of .4 m/s^2. A passenger arrives at a point next to the track 6.0s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train?

Homework Equations

$$\Delta$$x = v₀t + a/2*t²
V²_f=V²_i+ 2a*$$\Delta$$x

The Attempt at a Solution

I can find the position functions of both objects, but I can't rearrange the formulas just right where I'll end up with just one unknown.

i.e.

X_passenger = v₀t
X_train = 7.2 + 2.4*t + .2*t²

I know their positions have to be the same and I know their velocities have to be the same at that point (I picture it as a line touching the tip of a quadratic, like a derivative).

Since it's asking for a minimum, I'm also unsure of how to apply calculus. i.e. What am I taking the minimum of?

 

[Spinnor]

Homework Statement

A train pulls away from a station with constant acceleration of .4 m/s^2. A passenger arrives at a point next to the track 6.0s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train?

Homework Equations

$$\Delta$$x = v₀t + a/2*t²
V²_f=V²_i+ 2a*$$\Delta$$x

The Attempt at a Solution

I can find the position functions of both objects, but I can't rearrange the formulas just right where I'll end up with just one unknown.

i.e.

X_passenger = v₀t
X_train = 7.2 + 2.4*t + .2*t²

I know their positions have to be the same and I know their velocities have to be the same at that point (I picture it as a line touching the tip of a quadratic, like a derivative).

Since it's asking for a minimum, I'm also unsure of how to apply calculus. i.e. What am I taking the minimum of?

Also v(t) = v_o + a*t , v_o =0 so v(t=6) = ?

 

[zero_infinity]

Also v(t) = v_o + a*t , v_o =0 so v(t=6) = ?

Um, what?

The passenger is going to be moving with constant speed.

I know the answer - it's 4.8 m/s (i don't know how to arrive at it). That equation doesn't help me for the passenger, and I've calculated the velocity for the train (at t=6).

 

[zero_infinity]

Still need help with this. :/

Nobody knows how to do it? Or did I not follow the rules? I'm pretty sure I did..

 

[ehild]

The positions should be the same, the velocities do not.

_Xpassanger=x_train

This is an equation with two unknown, v₀ and t (the time needed for the passenger to reach the train).You need the minimum v₀.
It is possible to express v₀ as function of t and find t where v₀(t) is minimum.

ehild

 

[zero_infinity]

The positions should be the same, the velocities do not.

_Xpassanger=x_train

This is an equation with two unknown, v₀ and t (the time needed for the passenger to reach the train).You need the minimum v₀.
It is possible to express v₀ as function of t and find t where v₀(t) is minimum.

ehild

Yes, I have done that too. It's wrong

v(t) = .2(t^2 + 12*t + 36)/t

v'(t) = .2(t^2-36)/t^2 =0

this gives two extrema: 6 and -6

i already know the answer is 4.8, so... it doesn't work like that. I'm also pretty sure the velocities have to be the same.

http://img246.imageshack.us/img246/6751/posy.jpg

 

[ehild]

The time must be positive. Just substitute t=6 in v(t) = .2(t^2 + 12*t + 36)/t = .2(6^2 + 12*6 + 36)/6=? I got 4.8, but I might be wrong...

ehild

 

[zero_infinity]

The time must be positive. Just substitute t=6 in v(t) = .2(t^2 + 12*t + 36)/t = .2(6^2 + 12*6 + 36)/6=? I got 4.8, but I might be wrong...

ehild

How did you know they were going to meet at t=6(or t=12?)? It doesn't make any sense to me.

That's really not intuitive for me, wtf. If the passenger starts running at t=6, how they possibly meet at t=6(t=12?)?

 

[ehild]

"Xpassenger = v0t
Xtrain = 7.2 + 2.4*t + .2*t2"

Was it you who set up these equations? What does t mean?

ehild

 

[zero_infinity]

"Xpassenger = v0t
Xtrain = 7.2 + 2.4*t + .2*t2"

Was it you who set up these equations? What does t mean?

ehild

... time

the function for the train could have just been .2t^2(1/2*a*t^2), but i rewrote it for t=6 (found its position, velocity)

if you're plugging 6 into t, then it's 6 extra seconds on-top of the 6 after it started accelerating. how did you know to plug in 6?

 

[ehild]

Yes t is time but the time of what?
ehild

 

[zero_infinity]

what do you mean?!

in that equation it's the time it takes for them to meet at the same point

!?

 

[ehild]

You have set up two equations for the position of the passenger and for the position of the train


"Xpassenger = v₀t

Xtrain = 7.2 + 2.4*t + .2*t2"

and these are the same when they meet at t_m (meeting time):

v₀*t_m=7.2 + 2.4*t_m + .2*t_m^2 ---> v₀=7.2/t_m + 2.4+ .2*t_m.

The minimum value of v₀ is needed. You differentiated v₀ with respect to t_m and have found that it is zero for t_m =6 s, so the passenger needs the lowest speed if he caches the train in 6 s.

If you plug in 6 for t_m in the expession for v₀, you get 4.8 m/s.


v₀=7.2/6+ 2.4+ .2*6=1.2+2.4+1.2=4.8.

ehild

 

[zero_infinity]

You have set up two equations for the position of the passenger and for the position of the train


"Xpassenger = v₀t

Xtrain = 7.2 + 2.4*t + .2*t2"

and these are the same when they meet at t_m (meeting time):

v₀*t_m=7.2 + 2.4*t_m + .2*t_m^2 ---> v₀=7.2/t_m + 2.4+ .2*t_m.

The minimum value of v₀ is needed. You differentiated v₀ with respect to t_m and have found that it is zero for t_m =6 s, so the passenger needs the lowest speed if he caches the train in 6 s.

If you plug in 6 for t_m in the expession for v₀, you get 4.8 m/s.


v₀=7.2/6+ 2.4+ .2*6=1.2+2.4+1.2=4.8.

ehild

i'm an idiot

-_-

 

[ehild]

Well...