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 wit h a constant acceleration $\left(6.0\,\hat{i}+4.0\,\hat{j}\right)m /s^{2}$. The $x$ -coordinate of the particle at the instant when its $y$- coordinate is $32\, m$ is $D$ meters. The value of $D$ is :

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

Answer 1

Answer: (A)

$x = u_{x}t +\frac{1}{2}a_{x}t^{2}$
$y = u_{y}t +\frac{1}{2}a_{y}t^{2}$
$32 = 0\times t +\frac{1}{2}\left(4\right)\left(t\right)^{2}$
$t^{2} = 16$
$t = 4 \,sec
x = 3\times4+\frac{1}{2}\times6\times4^{2}$
$= 12 + 48 = 60 \,m$