An object has a velocity, `v = (2hati + 4hatj) ms^(-1)` at time `t = 0s`. It undergoes a constant acceleration `a = (hati - 3hatj)ms^(-2)` for 4s. Then
(i) Find the coordinates of the object if it is at origin at `t = 0`
(ii) Find the magnitude of its velocity at the end of 4s.

Options

A. `10 cm` , `8 m` , `rArr v = 6 hati - 8hatj`
B. `16 cm` , `18 m` , `rArr v = 6 hati - 8hatj`
C. `16 cm` , `8 m` , `rArr v = 2 hati - 8hatj`
D. `16 cm` , `8 m` , `rArr v = 6 hati - 8hatj`

Answer

D

Explanation

(i) Here initial position of the object
`r_(0) = x_(0) hati + y_(0) hatj +0 hati + 0hatj`
Initial velocity, `v_(0) = v_(0_(x))hati + v_(0_(y)) hatj = 2hati +4 hatj`
Acceleration `a = a_(x) hati +a_(y) hatj = hati - 3hatk`
And `t = 4s`
Let the final coordinates of the object be (x,y),. Then according to the equation for the path of particle under constant acceleration.
`x = x_(0) +v_(0_(x)) t +(1)/(2)a_(x)t^(2)`
`= 0 +2 xx 4 +(1)/(2) (1) xx 4^(2)`
`rArr x = 16 cm`
and `y = y_(0) +v_(0_(y)) t +(1)/(2)a_(y)t^(2)`
`= 0 + 4xx 4 +(1)/(2) (-3) xx 4^(2)`
`rarr y =- 8 m`
Therefore the object lies at `(16 hati - 8 hatj)` at `t = 4`s.
(ii) Using equation
`v = v_(0) +at`
`rArr v = (2hati + 4hatj) +(hati - 3hatj) xx 4`
`=(2hati +4hatj) +(4hati - 12 hatj) = (2+4) hati +(4-12) hatj`
`rArr v = 6 hati - 8hatj`
`:.` Magnitude of velocity, `|v| = sqrt(6^(2) +8^(2)) = 10 ms^(-1)`
Its direction with X-aixs `theta = tan^(-1) ((-8)/(6)) ~~ - 53^(@)`