Question

A particle starts from the origin at $t=0$ with an initial velocity of $3.0\hat{i}m/s$ and moves in the $x-y$ plane with a constant acceleration $(6.0\hat{i}+4.0\hat{j})m/s^2$. The $x-$ coordinate of the particle at the instant when its $y-$ coordinate is $32m$ is $D$ meters. The value of $D$ is

• A. $60$
• B. $32$
• C. $50$
• D. $40$

Answer 1

The correct option is A $60$

1. Find time $t$

Formula used : $s=ut+\frac{1}{2}a{t}^2$

Given, initial velocity is only in $x-$ direction,

$v_y=0$

And acceleration in $y-$ direction,

$a_y=4m/s^2$

And displacement in $y-$ direction,

$y=32m$

Using, $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$32=0+\frac{1}{2}\left(4\right)t^2$

$t=4s$

2. Find $D$.

Formula used : $s=ut+\frac{1}{2}a{t}^2$

Given, $x-$ coordinate of the particle at $4s$,

$=D$ meter

Initial velocity of the particle (along $x-$ axis),

$u_x=3m/s$

Initial acceleration of the particle (along $x-$ axis),

$a_y=6m/s^2$

Using, $x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$D=(3\times 4)+\frac{1}{2}\left(6\right)(4{)}^2$

$D=60m$

Final answer : (b)