A train stops at two stations A and B. It accelerates from rest from station A to a speed of 144 km/h^(-1) in 3 minutes and maintains this speed for 10 minutes. It then decelerates for 2 minutes and comes to rest at station B. Find the total distance between A and B.

Zaiden Soto

Zaiden Soto

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2022-08-19

A train stops at two stations A and B. It accelerates from rest from station A to a speed of 144 k m / h 1 in 3 minutes and maintains this speed for 10 minutes. It then decelerates for 2 minutes and comes to rest at station B. Find the total distance between A and B.

Answer & Explanation

afermavas4h

afermavas4h

Beginner2022-08-20Added 9 answers

Answer:
The total distance between A and B is 30 km
Explanation:
The given information are;
The time duration of the acceleration of the train from station A = 3 minutes = 0.05 hours
The speed attained by the train after the acceleration = 144 km/h
The time duration the train maintains the speed = 10 minutes = 0.1 hours
The time duration in which the train decelerates to station B = 2 minutes = 0.0 hours
The equation of motion required are;
The initial acceleration, a = ( 144 0 ) 0.05 = 2 , 880 k m / h 2
The distance covered, s = u t + 1 2 a t 2
Where;
u The initial velocity = 0
s 1 = 0 × 0.05 + 1 2 × 2880 × 0.05 2 = 3.6 k m
s 1 = 3.6 k m
The distance, s₂ the train covers at the constant speed 144 km/h for 10 minutes (1/6 hours) is given as follows;
s 2 =  Velocity Time  = 144 × 1 6 = 24 k m
s 2 = 24 k m
The deceleration, a₂ that brings the train to a stop in 2 minutes (1/30 hours) is given as follows;
a 2 = ( 0 144 ) ( 1 30 ) = 4320 k m / h 2
The distance covered, s₃ by the train as it decelerates to rest from the initial constant speed is given as follows;
s 3 = u t + 1 2 a t 2
Where;
u = The initial velocity =144 km/h
We have; s 3 = 144 × 1 30 1 2 × 4320 × ( 1 30 ) 2 = 2.4
s 3 = 2.4 k m
The total distance between A and B, s = s 1 + s 2 + s 3 = 3.6 + 24 + 2.4 = 30 k m
The total distance between A and B = 30 km.

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