Simple Kinematics

question 1:

let the two points be A and B.

now let velocity at A be `u`

vel at B= 45

using v=u+at , we have 45=u+6a ---(1)

using v²-u²=2as , we have

45²-u²=2.a.180 ---(2)

so u have 2 equations in 2 unknowns `u` and `a`.

solve to get` u` and `a`

good .. both are done at a single stroke..

1.
now.......v=45......s=180m....t=6s
v²-u²=2as
45*45-u²=2a.180................(1)
and also
v=u+at
45=u+6a
subs this in 1 equaton
(u+6a)²-u²=360a
u²+36a²+12ua-u²=360a
36a²+12ua=360a
36a+12u=360
u+3a=30
u+6a=45
3a=15
a=5m/s²

and u+15=30
u=15m/s..........

hehe b555.......shud i pink urs also............urs isnt complete though hehehehehehehe

Your first question answer is correct.Thanks !

The answer for the second one is 24m but i don't know how to get it.

tormented i havent really put it on a pepr but what akand has done seems to be correct.

i dont even hav a pen near me.............i was just typing.....so cud hav made sum mistake........

but will d retardation in both cases be same????? it has to be na......same car......

Yeah i don't think there is any fault in that but the book says 24m.By the way this question was a previous AIEEE question.

Okay i guess i got it !
s1 = 6 s2 = x
we know that s=u² / 2a and here 1/2 and a are constants so
s is proportional to u2
s1 / s2 = (u1)²/(u2)²

Solve and we get 24m !