Question

A particle is kept as a rest at origin. Another particle start from (5m, 0) with a velocity of -4i+3j m/s. Find their closest distance of approch​

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.