A particl leaves the orgin with an lintial veloity ` vec u = (3 hat I ) ms &(-1)` and a constant acceleration ` vec a= (-1.0 hat i-0 5 hat j)ms^(-1). Its velocity ` vec v` and positvion vector ` vec r` when it reaches its maximum x-coosrdinate aer .

Options

A. ` vec v= (- 3 hat ) ms^(-1)`
B. ` vec v = (- 2/2 hat j) ms^(-1)`
C. ` vec r = ( 3 hat - 2 hat j) m`
D. ` vec r = ( 9/2 hat I - 9/4 hat j) m`

Answer

B, D

Explanation

Here , ` u_x =3 ms^(-1) , a_x =- 1.0 ms^(-2)`,
u_y = 0 , a_y =- 0.5 ms^(-2_`
For maximum x-corrdinates, the final horizontal velocity becomes zero, ie. ` v-x =0`
As ` v_x^2 =u_x^2 + 2 a_xx` os `0 = 3^2 + 2(- 1) x`
or ` x= 4.5 m`
Also , ` v_x = u_x + a_x t` or =0 3 + (- 1) t`
or ` t= 3 s`
Taking motion along vettical direction, we have
` y=u_y t +1/2 a_y t^2 =0 3 + 1/2 (- 0.5)xx 3^2`
=- 2 . 25 m`
and ` v_y +a_y t=0 + (-0.5) 3 =- 1.5 ms^(-1)`
:. `vec v =v_x+v_y hat j =0 - 1.5 hat j = ( 1.5 hat j) ms^(-1)`
and ` vec r= xhat i+y hat j = (4.5 hat i - 2.25 hat j) m`.