Velocity of particle moving along positive x-direction is `v = (40-10t)m//s`. Here,t is in seconds. At time `t=0,` tha x coordinate of particle is zero. Find the time when the particle is at a distance of 60 m from origin.

Options

A. `t_3=2(1+sqrt7)s`
B. `t_3=2(2+sqrt7)s`
C. `t_3=2(2+sqrt9)s`
D. `t_3=sqrt7)s`

Answer

B

Explanation

Comparing with `v=u+at,` we have,
`u=40m//s` and `a=-10m//s^2`

Distance of 60m from origin may be at `x=+60m`
and `x=-60 m`.
From the figure, we can see that at these two points
particle is at three times `t_1,t_2` and `t_3.`
For `X=+60m` or `t_1` and `t_2`
`s=ut+1/2at^2`
`rArr 60=(+40)t+1/2(-10)t^2`
Solving this equation, we get
`t_1=2s` and `t_2=6s`
For `X-60m` or `t_3`
`s=ut+1/2at^2`
Solving this equation, we get positive value of t as
`t_3=2(2+sqrt7)s`