An express train passes through a station. It enters with an...

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 $\mathrm{m} / \mathrm{s}^{2}$ as it goes through. The station is 210 $\mathrm{m}$ long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) if the train is 130 $\mathrm{m}$ long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?

Key Concepts

Time Calculation under Uniform Acceleration

Time calculation in uniformly accelerated motion involves determining how long it takes for an object to travel a specific distance or reach a given velocity. By manipulating the kinematic equations, one can find time intervals that correspond to different stages of motion, such as when specific parts of the train pass a fixed point like the station boundary.

Displacement and Distance Relationships

This concept covers the total distance traveled by different parts of an object – in this case, different sections of the train. It is necessary to properly account for how distances such as the length of the station, the nose position, and the tail of the train correlate with one another during the deceleration process, helping to establish the relative timing of events like the nose or the tail exiting the station.

Constant Acceleration Kinematics

This key concept involves the use of kinematic equations that relate initial velocity, final velocity, acceleration (which in this case can be negative as deceleration), displacement, and time. These equations are essential in solving problems where an object moves with uniform acceleration, allowing us to determine unknown quantities such as time elapsed or final speed over a given distance.

Deceleration

Deceleration refers to a negative acceleration where the velocity of an object decreases over time. Understanding deceleration is crucial, as it affects the computation of how the speed evolves as the train travels through a station, and it is applied in the same set of kinematic equations by considering acceleration as a negative value.

Transcript

00:01 So here we're going to be using for part a, x final equals x initial plus vx initial times t plus one half a t squared.

00:10 Now here we can then say that plugging in for our values, we know that here there's no x initial and we're going to have to set up a quadratic equation.

00:23 So this term is eliminated and we then have that negative point zero.

00:33 75 meters per second squared times t squared plus vx initial t so plus 22 .0 meters per second times t and then minus the distance between train stations or we can say minus 210 meters this is equaling zero and this is essentially a quadratic equation.

01:13 We can use your solve function in your t -i -84, 85, or 89 in order to solve for t, and t equals now 9 .888 seconds and 283 seconds.

01:30 Now, we're not looking for the greater positive route because that represents the time when the train would come back to 210 meters if it moved backward after stopping.

01:43 So essentially we can eliminate this solution and the correct solution is 9 .88 seconds.

01:50 This would be our final answer for part a.

01:52 Now for part b, we are going to use the following expression.

02:00 We know that v final squared equals v initial squared plus two times the acceleration times the displacement in the x direction.

02:09 Again, we are going to then say that the final velocity would be equaling the square root of the initial squared plus two times the acceleration times the displacement in the x direction.

02:19 This would be equal to the square root.

02:21 The initial would be 22 .0 meters per second quantity squared plus two times the acceleration of negative 0 .150 meters per second squared times delta x of 210 meters.

02:42 And we find that then the final velocity is equaling 20 .5 meters per second.

02:49 This would be our final answer for part b for part c now we're going to suppose that in the beginning the train ends at position at the initial position so here the initial position is equaling negative 130 meters and this is allowed because the beginning of the station we're going to say is that position zero and then the end of the station 210 meters is positive so we can then say that here, x final is again 210 meters.

03:27 And x equaling zero meters, again, represents the beginning of the station.

03:32 So we can say that x final equals x initial plus vx initial times t plus one half times the acceleration in the x direction times t squared.

03:43 And here we're going to multiply by two and then we're going to have to rearrange.

03:48 So we have the a t squared plus 2 v x initial times t plus 2 times x initial minus 2 times x final and then this is going to equal 0...

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