Question

A car starts from rest with acceleration alpha and then retards to rest with retardation beta on a straight line such that total time of journey is T. the distance covered by the car during the journey is

Answer 1

let s1 and s2 be the distances covered
initially, u=0 and a = α then
from, V = U + at,
v1= αt1. now,
from, 3rd eqn of motion,we get,
(αt1)^2 = 0 + 2 α x s1
s1= (αt1^2)/2........(1)
SIMILARLY , now for second situation,
u=v1 , a = - β then,
v = u + at2
0 = αt1 - βt2
αt1 = βt2
or t1/t2 = β/α............(2)

from, v^2 = u^2 + 2 as
0 = v1^2 + 2(-β)s2
( αt1 )^2 = 2β(s2)
or s2 ={ ( βt2)^2}/2β from (αt1 = βt2)
s2= (βt2^2)/2
hence total distance covered ,
S = s1+s2
= (αt1^2)/2 + { βt2^2}/2
= (αt1^2)/2 + { βt2 × t2}/2 again (αt1 = βt2)
S = (αt1)/2 [ t1 + t2 ] SINCE( t1+t2=T)
S = [(αt1) T]/2