Question

A particle is kept at rest at origin. Another particle starts from (5m, 0) with a velocity of – 4i +3j m/s
Find their closest distance of approach.
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Answer 1

Answer:

ans should be 3m.........

After time t position of particle will be (5m−4t)iˆ+3tjˆ

So distance from the distance from the particle at origin is given by

 D = root under(5m−4t)^2+(3t)^2

Now for displacement to be minimum, 

dD/dt have to be zero, hence we have

d/dt root under(5m−4t)^2+(3t)^2 =0

⇒18t−8(5m−4t)/2×root under(5m−4t)^2+(3t)^2=0

⇒t=4/5

Now putting the value of t in the equation of distance we get, D=3. Hence the minimum of approach will be 3.