(a) If the maximum acceleration that is tolerable for passengers in a subway train is1.34m2 and subway stations are located 806mapart, what is maximum speed a subway train can attain between stations? (b)What is the travel time between stations? (c) If a subway train stops for localid="1660818170335" 20s at each station, what is the maximum average speed of the train, from one start up to the next? (d) Graph x, v and a versus t for the interval from one start-up to the next?

Short Answer

Expert verified

1.The maximum speed of a subway train that can attain between stations is 32.9ms.

2.The travel time between stations is 49.1s

3.The maximum average speed of the train from one start-up to the next11.7ms

Step by step solution

01

Given information

Acceleration of the train 1.34ms2

Distance between the stations 806m

02

Understanding the concept

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for third kinematic equation, the maximum speed of a subway train that can attain between the stations can be found. Further, using the formula for the second kinematic equation, the travel time between stations can be calculated.

Formula:

The final velocity is given by,

Vr2=V02+2ax

The displacement is given by,

x=v0t+12at2

Where V0the initial velocity, x is is the displacement.

03

(a) Determination of maximum speed attain by train between stations

Let’s consider the train accelerates from v0& x0at a1=1.34ms2 up to midway point

X1=8062m=403m

Where velocity is v1 and then train decelerates at a2=-1.34ms2 until it comes to a stop V0=0ms

So, maximum speed that train can attain is 32.9ms.

04

(b) Determination of travel time between stations

The time t1for the accelerating is

X1=V0t1+12a1t12

t1=2403m1.34ms2=24.53s

As time for decelerating is also same, total time t=2t1=49.1s is for travel between stations.

Therefore, the travel time between stations is 49.1s
.

05

(c) Determination of maximum average speed of the train from one start-up to the next

With dead time of 20stotal time from one stop to another stop is

T=t+20=69.1s

Vavg=80669.1m/s=11.7ms

06

(c) Graphs of  x, v and a versus t

Graph of distance vs time

Graph of velocity vs time

(iii) Graph of acceleration vs time

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