IIT JEE , AIEEE Forum - Physics , Mechanics

sneha_91

a car accelerates from rest at a constant rate "a" for some time and then deccelerates at a constant rate "b" to come to rest.if the total time taken is t sec,evaluate maximum velocity reached and the total distance travelled

timetraveller

Let it accelarate at rate 'a'for t' secs
so
max velocity = at'
now total time is 't'
and it decelarates at 'b' till rest

so=> 0= at' - b(t-t') [ v = u +at ]

=> t'(a - b) = bt

=> t' = bt/(a-b)

so max velocity => abt/(a-b)

a_buddy4uin06

let t be the time in which it attains max vel..
V= 0 + at (max speed)

V₂= V - b(T - t)
t = bT/(a - b)
max speed = abT/(a-b)
total displacement = ab²T² + ba²T²/ 2 ( a - b)²

elessar_iitkgp

This problem is best solved by a v-t graph

The slope of the first line is 'a' and that of the second is "-b"
Let the maximun velocity be v_m
Then, v_m/T = a

v_m/a = T ---------(1)
& v_m/t-T = b

v_m/b = t-T ---------(2)
Adding (1) and(2)
v_m = abt/(a+b)

As the v-t graph is entirely above the t axis,
So distance travelled = Displacement = Area under v-t curve = (1/2)v_mt
=abt² /2(a+b)

a_buddy4uin06

a small mistake yaar, the T( the one which we don kno), we hav to substitue the value tat we hav calculated.. so jus correct ur final ans...

a_buddy4uin06

hey sneha,, so our ans r rit na,...

elessar_iitkgp

Sneha has decided that she won't rate me!!!

sneha_91

i have rated u

elessar_iitkgp

cvramana

Thanks for the methonds shown. They are right!

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