Question

Three points A , B , C are on a straight line such Ab = BC . A particle moving acceleration passes A and C with speeds of 6m/s and 8 m/s. Another particle moving with uniform acceleration passes A and B with speed of 12m/s and 14m/s if t1 and t2 are time taken by the particle to cross AB t1/t2 is nearly?

Answer 1

The answer should be 2

Answer 2

time taken by the particles or particle?

Answer 3

It is not given we have to find it

Answer 4

Can anyone answer This!

Answer 5

@ajprincess can uanswer this

Answer 6

yeah, the answer is correct

Answer 7

Whuch answer?

Answer 8

the answer should be 2^^

Answer 9

u know the basic kinematics formulae?

Answer 10

yes

Answer 11

ok, say, the time taken are t1 and t2
and accelerations are a1, a2 respectively

Answer 12

ok

Answer 13

now, apply v^2=u^2+2as

Answer 14

as s is fixed, u can get the ratio of their acceleration

Answer 15

sorry, s is 2 times in one case^^

Answer 16

can u get the ratio?

Answer 17

I gave Up !! NO Idea

Answer 18

i am disappointed.. u r not trying

Answer 19

i will try my best

Answer 20

say, the first person's acceleration and time are a1, t1
second person's acceleration and time are a2, t2
say, AB=s
then applying the kinematics formulae..
8^2=6^2+2a1*2s
or s=7/a1

14^2=12^2+2a2*s
or s=26/a2

now, equate s from both cases
7/a1=26/a2
a2/a1=26/7

got it so far?

Answer 21

i got a2=13a/7

Answer 22

no, look carefully, first person covers AC, second person covers AB
so, first person goes 2s, second one goes s

Answer 23

in question it is said that AB = Bc

Answer 24

@Arnab09 so diatances are same

Answer 25

so, AC=2AB

Answer 26

yeah u r correct!

Answer 27

Plz go on !

Answer 28

okay, then apply:
v=u+at

so, 8=6+a1*t1

t1=2/a1

14=12+a2*t2

so, t2= 2/a2

so, t1/t2= a2/a1
as proved before, a2/a1= 26/7
so, t1/t2= 26/7
answer will be 4 approx

got it?

Answer 29

@Arnab09 the answer should be 2

Answer 30

@mahmit2012 i think u r correct