Question

two motor cars start off with a gap of 2 sec with the same acceleration. How long after the departure, does the second car travel a distance equal to 1/9 th the distance covered by the first car

Answer 1

S1=1/2at1^2
1/9S1=1/2at2^2
S1=9/2at2^2
1/2at1^2=9/2at2^2
t2=t1/3
t-2=t/3
2t=6
t=3
the second car will take 1/3 rd of the time taken by the first car to cover 1/9 th of distance.
I think this is.......the ans......1second
sorry if wrong..........

Answer 2

Relation used : s = ut + 1/2*a*t^2
It is assumed that both cars starts from rest hence initial velocity is zero for both cars
Let 't' be the time at which the above situation occurs
'a' be the common acceleration
Let s1 = distance travelled by car 1
Let s2 = distance travelled by car 2

therefore, s2 = 1/2*a*t^2 ---- (1)
s1 = 1/2*a*(t+2)^2---- (2)

as per the condition given ,
s2 = s1/9

substituting the value of s2 in (1) and dividing (2) by (1) we get,
9 = ((t+2)/t)^2

taking square roots we get ,
3= t+2/t

which gives t = 1 sec