IIT JEE , AIEEE Forum - Physics , Mechanics

A.Vignesh

Plz try this:

The motors of an electric train can give it an acceleration of 1 m/sec² and the breaks can give a negative acceleration of 3 m/sec² . the shortest time in which the train can make a trip between two stations 1215 m apart is

shubhayan_c

I think the answer is around 56 seconds.
i've tried it in my copy. if u want den i'll giv the detailed soln.

A.Vignesh

yeah pls post the detailed solution

shubhayan_c

try making it's v-t graph and the area under it vil b the answer.

rishikesh_anshu

if we draw v-t graph for min time velocity first increses with slope 1 and achieve max velocity v then slows down with slope 3
total area = 1/2vt=1215
let accelating time = t1=v
retarding time = t2=v/3
t=t1+t2=4v/3
puttin in eq 1
(1/2)(4v/3)v=1215
v=root(1215*3/2)=42.69
t=4v/3=56.92

shubhayan_c

Good effort i was using intergration for finding the area dat's why i think i didn't got the exact answer.

arvind1990

This is a shortcut method!!!

s=1/2(ar/a+r)t² This is the general formula!!!

1/2(1x3/1+3)t²=1215!!which is 56.92099...!!!

Rate me!!!

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