Question

A point moving with const. acc. from A to B, In a straight line AB has velocities u and v at A and B respectively.The velocity of point at C,the mid point of AB is?

Answer 1

@mukushla :P

Answer 2

got it

Answer 3

first find the distance AB using
v^2 = u^2 + 2as
s = (v^2 - u^2) / 2a = AB
now we want the final velocity when displacement is s/2
substitute in the eq V^2 = u^2 + 2as
and u'll get V = sqrt [ (u^2 + v^2)/2 ]

Answer 4

@Vaidehi09
nice

Answer 5

^^

Answer 6

ok m gonna take velocity at C as L.
so, L^2 = u^2 + 2a ( s/2)
L^2 = u^2 + 2a [ (v^2 - u^2) / 4a ]
= u^2 + (v^2 - u^2)/2
= [ 2u^2 + v^2 - u^2 ] / 2
L^2 = [ u^2 + v^2 ] /2
so L = sqrt ( [u^2 + v^2]/2 )

Answer 7

got it?

Answer 8

got it thanks ! :)

Answer 9

gr8!

Answer 10

in this ques o.o if time from A to C is twice that from C to B then which of the following relation is correct?
v=4u
v=7u
v=8u
v=11u

Answer 11

i tried v=u+at,ended up nowhere
also the answer is v=7u :)

Answer 12

i think i got it

Answer 13

just give me a hint. not the whole sol.

Answer 14

\[t_{AC}=\frac{2}{3} t_{AB}\]

Answer 15

and
\[v_{c}=\sqrt{\frac{u^2+v^2}{2}} \]

Answer 16

i got v^2 +7u^2 -8uv = 0

Answer 17

v^2 +7u^2 -8uv =(v-7u)(v-u)

Answer 18

oh yea!...ok got it!

Answer 19

thanks :))

Answer 20

thank u too.