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.

Explanation

(i) Here original position of the object.
` vecr_(0) = x_(0)hati + y_(0)hatj = 0 hati + 0 hatj`
Initial veclocity `vecv_(0) = v_(0)x hati + v_(0y) hatj = 2hatj + 4 hatj`
`veca = a_(x)hati + a_(y)hatj = hatj - 3hatj`
And t = 4 s.
Let the final co-rodinates of the object be (x,y) . then according to the equation (iii) derived in previous section.
` x = x_(0) + v_(0x)t + 1/2a_(x)t^(2)`
` = 0 +2 xx 4 + 1/2 (1) xx 4^(2)`
x = 16
`and y = y_(0) +v_(0y)t + 1/2 a_(y)t^(2) = 0 + 4 xx 4 + 1/2(-3) xx 4^(2)`
y = -8
Therefore the object lies. at (16 -8) at t= 4s.
(ii) Using eqution.
`vecv = vecv_(0) + vecat`
` Rightarrow vecv = (2hati + 4hatj) + ( hati - 3hatj) xx 4`
` = (2hatj + 4hatj) + ( 4hatj -12 hatj)`
` = (2+4 ) hatj + ( 4 -12)hatj`
` Rightarrow vecv = 6hati - 8o hatj ` : velocity at the end of 4s.
` |vecv| = sqrt(6^(2) +8^(2)) = 10 ms ^(-1)`
Its direcationn with x-axis, ` theta = tan^(-1) (-8)/6 = 53.13^(@)`