A force of `(2.6 hat(i) + 1.6 hat(j))` N acts on a body of mass 2 kg. If the velocity of the body at time, t = 0 is `(3.6 hat(i) - 4.8 hat(j)) ms^(-1)`, the time at which the body will just have a velocity along x-axis only is

Options

A. 1s
B. 2s
C. 3s
D. 6s

Answer

D

Explanation

Given, force acts on a body,
F = `(2.6 hat(i) + 1.6 hat(j))` N
mass of a body , m = 2kg
At t = 0 , velocity of the body, v = `(3.6 hat(i) - 4.8 hat(j) ) ` m/s
we know that,
Force = mass `xx` acceleration
`therefore " " F = m xx a `
or `" " a = (F)/(m)`
or `" "a = ((2.6 hat(i) + 1.6 hat(j))/(2)) `
or `" " a = (1.3 hat(i) + 0.8 hat(j)) m//s^(2)`
now, the velocity vector, `(dv)/(dt) = a`
`int dv = int a` dt
or v = `(1.3 hat(i) + 0.8 hat(j)) t + c`
at t = 0 ,c = v = `(3.6 hat(i) - 4.8 hat(j) ) ms^(-1)`
`therefore v = (3.6 + 1.3 t) hat(i) + (-4.8 + 0.8t) hat(j)` m/s
`therefore v_(y) = 0` (because body will just have a velocity along x - axis. )
-4.8 + 0.8t = 0
`rArr " "` 0.8t = 4.8
`rArr " " t = (4.8)/(0.8) = 6s `
This is the time at which the body will just have velocity along x- axis
t = 6 s