(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m /

(a) If the maximum acceleration that is tolerable for...

Key Concepts

Constant Acceleration Kinematics

This concept involves using the kinematic equations, which relate displacement, velocity, acceleration, and time. When acceleration is constant, these equations allow us to calculate one variable if the others are known. It is a fundamental principle in physics that is used to analyze the motion of objects, such as subway trains, over time.

Symmetrical Acceleration and Deceleration

In many motion problems, especially in transit systems, the journey is divided into two phases: acceleration to a maximum speed and then deceleration back to a stop. Assuming the magnitudes of acceleration and deceleration are equal, the analysis can be simplified by treating the two halves of the journey as mirror images, enabling the determination of maximum speed from the distance available.

Distance Partitioning in Motion

This concept refers to dividing the total distance traveled into segments where different motion parameters apply. In the context of a subway train, the available distance between stations is partitioned into the portion needed for acceleration and the portion required for deceleration. This partitioning is essential to optimize travel by determining the maximum attainable speed within the constraints given.

Average Speed Calculation with Dwell Time

In transportation scenarios, the effective or average speed must account not only for the time spent moving but also for the time spent stopped, such as at station dwell times. Calculating the average speed over an entire trip—including both movement and stoppage periods—is crucial for assessing overall performance and efficiency in transit schedules.

Graphical Analysis of Motion

Graphing displacement, velocity, and acceleration as functions of time provides a visual representation of motion, which aids in understanding the dynamics of acceleration and deceleration. Interpreting these graphs helps in the analysis of how speed changes over time, the impact of acceleration limits, and the cumulative displacement achieved, all of which are key in planning efficient transit operations.

Transcript

00:01 In this problem of motion in a straight line we have given that if the maximum acceleration that is tolerable for passenger in a subway is so this is the maximum acceleration that is equal to 1 .34 meters per second square so this is the maximum tolerable acceleration and subway stations are located 806 meters apart so let's suppose this is the we can say this is station 1 -2 station 2 say we are naming it as station a and then station b and the distance between station 1 to station 2 is equal to 8606 meters what is the maximum speed a subway train can attain between the station so let's suppose at station 1 this would be at 0 we can say the initial velocity would be 0 and position is also 0 and from here it attains the maximum acceleration this is equal to 1 .34 meters per second square then until the midway point and then say this is the midway point point c and then we have to decelerate it with the same maximum acceleration so this is minus 1 .34 meters per second square and here we have plus 1 .34 meters per second square that means we have to find the maximum velocity as point c or we can say the velocity at point c so this would be the maximum velocity so at midpoint ac would be equals to 806 divide width 2 which is equal to 403 meters now we have the formula that is v1 this is square is equal to v0 square plus twice the acceleration which is equal to 1 .34 meters per second is square and multiplied with say this is the distance which is equal to 403 meters now this is equal to 0 and v1 is under root of this term which is 2 multiplied with 1 .34 multiplied with 403 now when we solve this under root v1 is equal to 32 .9 meters per second so this is the answer for part a so we can say the answer for part a is this is the maximum velocity which is equal to 32 .9 meters per second now for part b part b says what is the travel time between these stations so we have to find the travel time so this is x1 is equals to v0 t plus half multiplied with acceleration say a1 and t1 square so from here we say that v0 is equal to 0 so this is say this is t1 and this term would be equal to 0 and now we have to find t1 so t1 is equal to n under root of now we are finding the time taken from a to c so this value would be twice of x1 which is equal to 403 divided with acceleration which is 1 .34 which is equals to 24 .53 seconds from here we say that the time taken from a to c is equal to 24 .53 and now from c to b we have the identical acceleration and identical distance of time taken would be twice of t1.

03:41 So twice of t1 is equals to this is 49 .1 seconds.

03:47 So for part b we have the time which is equal to 49 .1 seconds.

03:55 So this is the answer for part b.

03:57 Now in part c we have given that if a subway train stops for 20 seconds at each station what is the maximum average speed of the train for from one start up to the next.

04:14 That means at point a this is 20 second halt and at point b we have again 20 second halt.

04:24 So let's suppose our train comes from any of the previous station say this was at station x.

04:31 Then when it comes to a we have to count down the time.

04:35 So this would be resting for 20 seconds and then it would be continued for motion until it reaches for b...

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