Constant acceleration problem with two objects

[Portishead]

Homework Statement

A train pulls away from a station with a constant acceleration of 0.40 m/s^2. A passenger arrives at a point next to the track 6.0 s 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? On a single graph, plot the position versus time curves for both the train and the passenger.

Homework Equations

x-train(t)=.20t^2
x-passenger(t)=vt+x_o

The Attempt at a Solution

I initially supposed that x-train=x-passenger at t=6.0 s but that's clearly wrong. The answer at the back of the book is that v=4.8 m/s but I'm not sure why.

 

[w3390]

Does the passenger need to catch up to the train before a certain distance, like before the end of the train station?

 

[Portishead]

Does the passenger need to catch up to the train before a certain distance, like before the end of the train station?

I don't think so, as it is not given.

 

[w3390]

Okay. So you need to be careful when setting up your equations here. I will give you a hint: The position of the passenger at time t is equal to the position of the train at time (t-6).

 

[Portishead]

So, wouldn't it be the other way around? Train(t)=.20t^2; Passenger(t)=v(t-6)

 

[rl.bhat]

When passenger reaches the starting point, the velocity of train is

u = at and xo = 1/2*at^2.

If v is the velocity of the passenger, in time t1 distance traveled by her to catch the train is x = v*t1.

During that time the distance traveled by the train is = x1 = u*t1 + 1.2*a*t1^2

Now x = xo + x1.

v*t1 = 1/2*a*t^2 + a*t*t1 + 1/2*a*t1^2

v = (1/2*a*t^2)/t1 + (a*t*t1)/t1 + (1/2*a*t1^2)/t1

Are you familiar with the differentiation?

If yes, find differentiation of v with respect to t1 and equate it to zero to find the minimum velocity.

 

[Portishead]

rl.blat, can you take an ordinary derivative of that expression? t and t1 are different variables? Also, uh I solved the problem. I minimized v-passenger=x-train(t')-x-passenger(6)/(t'-6)=.20(t')^2/(t'-6) where t' is the time she gets on the train. That function has a relative minimum at t'=12 s if I did my calculus right.

 

[rl.bhat]

Your answer is correct. In my solution t is constant and is equal to 6 s. There are only two variables, v and t1.