A train which stops at all stations travels a total distance of 2.0 km between t

Question

A train which stops at all stations travels a total distance of 2.0 km between two consecutive stations assuming it to accelerate from the first station at a constant rate of 1.3 ms -2 for the first 300 m of the trip maintain constant speed for the next 1200 m and decelerate uniformly for the final distance to the 2nd station

A. The maximum speed attained by the train ?
B. the acceleration of the train during the final distance to the second train
C. The time taken to Tavel between the two stations

Explanation / Answer

a) it accelerates over the 300 m so we use v^2 = v0^2 + 2 a d v^2 = 0^2 + 2 * 1.3 * 300 v=27.9 m/s b) it went 300 and then 1200 for a total of 1500 m so 500 m left to go once again we use v^2 =v0^2 + 2ad 0 = 27.9^2 + 2*a*500 a= -0.778 m/s^2 c) time to accelerate x=x0+v0x t + 1/2 at^2 300 = 0 + 0t + 0.5*1.3*t^2 t=21.48 time at constant speed 1200 = 0 + 27.9 t + 0 t= 43.01 time to deaccelerate 500 = 0 + 27.9 t +0.5*-0.778t^2 t=35.04 s total time = 21.48+43.01+35.04 = 99.53

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