A train pulls away from a station with a constant acceleration of 0.44 m/s2. A passenger arrives at a point next to the track 6.2 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?

distance train=1/2 a t^2

distance passenger=V*(t-6)

set them equal

1/2 a t^2=Vt-6V

.22t^2-Vt+6V=0

t=(V+-sqrt(V^2-.88*6V)/.44

Well, V^2>.88*6V
V>5.20m/s

Lets see if it works.

t=(5.20+-sqrt(0))/.44=12 seconds

distance train: 1/2 .44*144=you do it
distance passenger= 5.20(12-6)=you do it

The response by @bobpursley seems incorrect. I had a different method of solving for Vo for the passenger train.

To answer this question, we need to consider the motion of the train and the motion of the passenger.

Let's analyze the motion of the train first. We have the acceleration of the train, which is 0.44 m/s^2. Since the acceleration is constant, we can use the equation of motion:

d = v0t + (1/2)at^2

where d is the distance covered, v0 is the initial velocity, t is the time, and a is the acceleration. In this case, the initial velocity (v0) is assumed to be zero because the train starts from rest.

Now, at the point when the passenger arrives, the distance traveled by the train is the same as the distance between the passenger and the starting point of the train. We can set up an equation to find this distance:

d = 0.5at^2

where d is the distance traveled by the train (from the starting point of the train) and t is the time it takes for the passenger to reach that point after the end of the train has passed.

Given that t = 6.2 s, we can solve for d:

d = 0.5 * 0.44 * (6.2)^2

d ≈ 8.72 meters

Now, let's consider the motion of the passenger. The passenger needs to cover this distance in the same amount of time it took the train to reach that point. Since the passenger starts from rest as well, we can use the equation of motion:

d = v0t + (1/2)at^2

where d is the distance covered, v0 is the initial velocity (which is the speed the passenger needs to run at), t is the time, and a is the acceleration.

Substituting the values we have, the equation becomes:

8.72 = (1/2) * v0 * (6.2)^2

Now we can solve for v0, the slowest constant speed at which the passenger can run and still catch the train.

no