IIT JEE , AIEEE Forum - Physics , Mechanics

blonkers

2 particles are initially located at points P & Q a distance D apart.They start moving at time t =0 such that the velocity u of Q is always along the horizontal

direction and the vlocity v of P is continually aimed at Q find time taken for the 2 particles to meet

sahilgupta_iit

it means that P always moves towards Q. in that case the time becomes d/v+u

dream

is the answer

D/squareroot(v^2-u^2)

blonkers

initially the velocity is perpendicular to u

rathin_123

if they both do constant acc motion then &their acc is A

then time t=root(D / A)

for p it meet q at D/2 as they both do con acc

THEIR u=0 so from D/2=ut+1/2 At^2

D/2=1/2 At^2

t=root(D / A)

blonkers

the velocity of Q is always constant
but the velocity of P is no as it is always changing direction

rathin_123

i didn't mantion they have same velocity.they do con acc.

and hey....

p&q both moves to each oyher as you told.

blonkers

what i meant was Q is moving along the horizontal
to be precise suppose Q is a point (0,D) then P is (0,0)
Q is moving parallel to x-axis with a constant velocity and P is moving with a velocity v but the direction is towards Q

utkarshsingh_91

if Q is moving towards opposite direction of P .. then time = (d/(u-v))
else
if Q is moving towards P..then time = (d/(u+v))

blonkers

P is moving towards Q

elessar_iitkgp

Let at any instant, the velocity vector of P makes an angle with the direction of motion of AQ.

Then, the velocity of approach

-(dl/dt) = v - u cos

-_d⁰ dl = vT - u₀^T cos dt

d = vT -u ₀^T cos dt ------------(1)

After sufficient time B moves on the same line as B, only separated by a distance, say x.

uT =v ₀^T cos dt

₀^T cos dt = uT/v ---------------(2)

From (1) and (2)

x=dv/(v²-u²)

satmat_007

let they meet at a distance x from Q.then

x/u=d+x/v (they meet at d same time)

solving gives x=ud/v-u
t=d/v-u

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