Question

Two particles P and Q move in the xy plane. Find the time where they are closest, their shortest distance at this point. At t = 0, P starts from point 8j and moving with speed 5i + j. At t = 0.5 s, Q starts from point 17i + 2h and moving at a speed of 2i + 4h​

Answer 1

Answer:

The distance between them first decreases and then increases

C

The shortest distance between them is 2

2

m

D

Time after which they are at minimum distance is 1s

V

AB

=

V

A

−

V

B

=2(

i

−

j

)⇒∣

V

AB

∣=2

2

.

Assuming B to be at rest, A will move with velocity

V

AB

in the direction shown in figure. The distance between them will first decrease from A to C and then increase beyond C.

Minimum distance between them is BC which is equal to

2

4

or 2

2

and the time after which they are at closest distance is :

t=

V

AB

AC

=

2

2

2

2

=1second.