SOLUTION: . You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceleration of 0.600 m/s^2. You arrive at the station ex

Question 819667: . You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceleration of 0.600 m/s^2. You arrive at the station exactly 4.00 seconds after the train leaves and you immediately start running after the train with a constant velocity of 8.50 m/s.
a. How long after the train leaves the station do you catch up with the train? 4.82 s and 23.5 s
b. How far from the train station do you catch up with the train? 6.97 m and 166 m
c. With what minimum speed would you have to run in order to catch up with the train? 4.8 m/s

Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
For an object moving in a straight line with a constant acceleration, you can use the average velocity to calculate the distance traveled.
with
= velocity (a function of time)
= initial velocity (at time = zero)
= time
= acceleration
Here , so the velocity as a function of time is
and the average velocity is

So, for the train, the distance traveled, , as a function of time since leaving the station is

a. Running at a constant speed of 8.50 m/s, the distance run as a function of time since the train left is

When you catch the train, you will have run the same distance that the train has traveled since leaving the station, so

The solutions to that equation can be found using the quadratic formula

You get the two solutions and .
That means that you catch up with the train only 4.82 seconds after the train left the station. That means that you catch up to the train after running for only 0.82 seconds. It is not surprising because the train accelerated at a low rate and it was going very slowly.
The second result is the silly one. If you keep running at that speed, you will pass the slow moving train, but the train continues accelerating, and 23.5 seconds after leaving the station, it will catch up to you. By then, very sweaty and tired, you could jump on the fast-moving train.

b. After running for 0.82 seconds you covered
(8.50m/s)(0.82s)=6.97m (Let's say 7 meters).
If you wanted to keep running past the train and wait for the train to catch up to you, 23.5 minutes after if left the station, you would have to run a distance, in meters, of
(Let's say 166 meters).
However, at that point the train will be going at a velocity in meters per second of

It may not be pleasant to jump into a train going that much faster than you are running.
Here is the graph of distance from the station as a function of time since the train left:
The train curve is the red parabola; you are represented by the green line.

c. The minimum speed you would have to run in order to catch up with the train?
is the speed (the slope of a line that makes it tangent to the train curve (with the line staring from 4 at the time axis.
That is the speed that would make the quadratic equation have just one solution.
The equation would be

For that equation to have just one solution we need

The solutions are , which does not make sense,
and .
So you need to be able to run at least 4.8 m/s.
In that case, the equation is

(dividing by )
and the solution is ,
meaning that you would catch the train 8 seconds after it left the station, after running for 4 seconds at 4.8 m/s.
To boot, the curves are tangent, meaning that you and the train are going at the same velocity when you meet so it is a smooth transfer.
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