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A train starts from a station P with a uniform acceleration a1, for some distance and then goes with uniform retardation a2 for some more distance to come to rest at the station Q. The distance between the stations P and Q is 4Km and the train takes 4 minutes to complete this journey, then 1/a11/a2=

A train starts from a station P with a uniform acceleration a1, for some distance and then goes with uniform retardation a2 for some more distance to come to rest at the station Q. The distance between the stations P and Q is 4Km and the train takes 4 minutes to complete this journey, then 1/a11/a2=

### Asked by Petulla Mishra | 4th Nov, 2015, 10:10: AM

Expert Answer:

###
There are two case in given question:
1. When train moves with uniform acceleration
2. When train moves with uniform retardation.
For the 1st case,
Let the train start from a station P with uniform accleration a_{1}, with initial velocity u_{1} and t_{1} be the time when it reaches at some distance s_{1}
From kinematic eqution of motion we have,
v_{1}=u_{1}+a_{1}t_{1 }(u_{1}=0 Initialy at rest)
v_{1}=a_{1}t_{1 .... }(i)
For the 2nd case the final velocity of first interval will be intial velocity of second.
Let a_{2} be uniform retardation, t_{2} be the time when it travels a distance s_{2} and comes to rest .
From kinematic equation of motion we have,
v_{2 }= u_{2 }- a_{2}t_{2 }= v_{1 }- a_{2}t_{2} (v_{2 }= 0 finaly comes at rest)
v_{1 }= a_{2}t_{2 ... (iii)}
Using kinematic equation of motion,

There are two case in given question:

1. When train moves with uniform acceleration

2. When train moves with uniform retardation.

For the 1st case,

Let the train start from a station P with uniform accleration a

_{1}, with initial velocity u_{1}and t_{1}be the time when it reaches at some distance s_{1}From kinematic eqution of motion we have,

v

_{1}=u_{1}+a_{1}t_{1 }(u_{1}=0 Initialy at rest)v

_{1}=a_{1}t_{1 .... }(i)For the 2nd case the final velocity of first interval will be intial velocity of second.

Let a

_{2}be uniform retardation, t_{2}be the time when it travels a distance s_{2}and comes to rest .From kinematic equation of motion we have,

v

_{2 }= u_{2 }- a_{2}t_{2 }= v_{1 }- a_{2}t_{2}(v_{2 }= 0 finaly comes at rest)v

_{1 }= a_{2}t_{2 ... (iii)}Using kinematic equation of motion,

### Answered by Faiza Lambe | 4th Nov, 2015, 11:06: AM

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