Let `r_(1)(t)=3t hat(i)+4t^(2)hat(j)`
and `r_(2)(t)=4t^(2) hat(i)+3that(j)`
represent the positions of particles 1 and 2, respectiely, as function of time t, `r_(1)(t)` and `r_(2)(t)` are in metre and t in second. The relative speed of the two particle at the instant t = 1s, will be

Options

A. 1m/s
B. `3sqrt(2)m//s`
C. `5sqrt(2)m//s`
D. `7sqrt(2) m//s`

Answer

C

Explanation

Here,
`vecr_(1)(t)=3thati+4t^(2)hatj`
`vecr_(2)(t)=4t^(2)hati+3thatj`
velocity, `vecv_(1)(t)=(dvecr_(1))/(dt)=(d)/(dt)(3t hati+4t^(2)hatj)`
`=3hati+8hatj`
`vecv_(2)(t)=(dvecr_(2))/(dt)=(d)/(dt)(4t^(2)hati+3t hat j)=8t hati+3 hatj`
The relative speed of particle 1 with respect to particle 2 is
`vecv_(12)=vecv_(1)-vecv_(2)`
`=(3hati+8thatj)-(8thati+3hatj)`
`=(3-8t)hati+(8t-3)hatj`
At t=1s,
`vecv_(12)=(3-8)hati+(8-3)hatj`
`=-5hati+5hatj`
`|vecv_(12)|sqrt((-5)^(2)+(5)^(2))`
`=sqrt(25+25)=5sqrt(2) m//s`