Question

A train starts from rest and moves with a constant acceleration on 2.0 m/s for half a minute. The brakes
are then applied and the train comes to rest in one minute after applying breaks. Find fa) the total distance
moved by the train, JK) the maximum speed attained by the train and (y the position(s) of the train
at half the maximum speed. (Assume retardation to be constant)​

Answer 1

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Answer 2

Answer:

• Total distance moved by train = 6300 m
• Maximum speed attained by train = 60 m/s
• Position of Train at half the maximum speed = 2250 m from the starting point.

Explanation:

Firstly

• Initial velocity of train = 0
• Constant acceleration of train = 2.0 m/s²
• time for which train moves with this this acceleration = 30 sec

So, calculating final velocity of Train after 30 sec

Using first equation of motion

→ v = u + a t

→ final velocity after 30 sec = 0 + 2 ( 30 )

→ final velocity after 30 sec = 60 m/s

Now, since after 30 sec Brakes are applied therefore,

• Maximum speed attained by the train = 60 m/s.

Now, calculating distance covered by Train in first 30 sec

Using second equation of motion

→ s = u t + 1/2 a t²

→ distance covered in first 30 sec = ( 0 ) ( 30 ) + 1/2 ( 2 ) ( 30 )²

→ distance covered in first 30 sec = 900 m

Further As given that,

After 30 sec brakes are applied and train comes to rest in one minute means 60 sec after applying brakes.

so, in this case initial velocity = 60 m/s

final velocity = 0

Hence, calculating deceleration of train

Using first equation of motion

→ 0 = 60 + a ( 60 )

→ a = -1 m/s²

Now, Calculating distance covered by train in last 1 min

Using second equation of motion

→ distance covered in last 60 sec = ( 60 ) ( 60 ) + 1/2 ( 1 ) ( 60 )²

→ distance covered in last 60 sec = 3600 + 1800

→ Distance covered in last 60 sec = 5400 m

So, now we can calculate the total distance covered by Train

→ Total distance covered = ( distance covered in first 30 sec ) + ( distance covered in last 60 sec )

→ Total distance covered = 900 + 5400

→ Total distance covered = 6300 m

Therefore,

• Total distance moved by the train = 6300 m.

Further, we need to find the position of Train at half the maximum speed.

[ taking retardation constant ]

half of the maximum speed will be 60/2 = 30 m/s

so, Using third equation of motion

→ 2 a s = v² - u²

→ 2 ( -1 ) (distance) = ( 30 )² - ( 60 )²

→ distance = ( 900 - 3600 ) / -2

→ distance = 1350 m

Therefore,

• position of train at half the maximum speed would be; ( 900 + 1350 ) m = 2250 m from the starting point of train.