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Grade: 12th pass
        
A train starts from rest from a station with acceleration 0.2 on a stright line and then comes to rest after attaining maximum speed on another station due to retardation 0.4. if total time spent is half an hour, then distance between two stations is?
4 years ago

Answers : (1)

Paarth Arkadi
22 Points
							
Since the units of accleration / retardation are not given i assume it to be km/h2.
This is a case of 1-D motion.
Let the accleration and retardation of the train be \alpha and \beta respectively.
The train starts from rest and attains a velocity vmax then it starts retarding and finally comes to rest.
Hece the motion can be divided into two parts accleration and retardation.
Taking the first part,
u (initial velocity) = 0                  …. ( since train starts from rest)
v = vmax                                         …..( final velocity)
time = t1                                         …..( total time = t1 + t2 )
By using the equation,
v = u + at
vmax = 0 + \alphat1
Vmax \alphat1    …......(1)
Similarly,
Taking second part,
0 = vmax \betat2
-vmax = -\betat2
vmax \betat2  …...........(2)
From eq (1) and (2),
t= and  t= vmax / \beta
 
Total time = t = t+ t2 = vmax /\alpha  + vmax / \beta   
Therefore , t = \beta vmax \alpha vmax \alpha\beta …................(3)
Since it is given that total time (t) = 0.5 hrs
Substituting the values in equation (3) we get,
0.5 = 0.4 vmax + 0.2 vmax / 0.4*0.2
vmax = 0.04
When we plot the velocity – time graph we get a traingle with its peak as vmax ( that is height ) and base as 0.5
Since the area under velocity – time graph gives displacement ( in this case same as distance ) we have,
distance / displacement = ½ * vmax * total time (t)
                                           = ½ * 0.04 * 0.5
                                           = 0.01 km or 10 m
 
4 years ago
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