Question

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 2.15 m/s² for 15.4 s/ It then proceeds at constant speed for 1180 m before slowing down at 2.2 m/s² until it stops at the station. What is the distance between the stations?

Answer 1

Total distance =Distance covered while accelerating +1180 m +Distance covered while decelerating

Acceleration

initial velocity, u =0m/s

acceleration, a =2.15 m/s²

time, t =15.4s

Use Formula $s=ut+1/2at^{2}$

$s_{1}=0m/s*15.4s+0.5*2.15m/s^{2}*(15.4s)^{2}$

$s_{1}=0+0.5*2.15*15.4^{2}$

$s_{1}=254.947m$

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Now find the constant speed attained

Use Formula $v=u+at$

$v=0m/s+2.15m/s^{2}*15.4s$

$v=33.11m/s$

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Now distance covered while decelerating

Final velocity is zero

Use Formula $v^{2}-u^{2}=2as$

$(0m/s)^{2}-(33.11m/s)^{2}=-2*2.2m/s^{2}*s_{3}$

$0^{2}-33.11^{2}=-2*2.2*s_{3}$

$33.11^{2}/(2*2.2)=s_{3}$

$s_{3}=249.153$

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Total distance =$s_{1}+1180m +s_{3} =254.947m+1180m+249.153m$

ANSWER: $1684.1m$

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