Question

A particle is given velocity $3m/s$ along the positive $x-\text{axis}$ subjected to constant acceleration $4m/s^2$ along the positive $y-\text{axis}$. Find the displacement (in $m$) of the particle in $2s$.

Answer 1

Draw a labelled diagram.

Find the displacement.

In x: (Uniform motion)

$x=u_{x}t=3\times 2=6m$

In $y$: (constant acceleration)

Therefore, we use
$s=ut+\frac{1}{2}a{t}^2$
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$y=0+\frac{1}{2}\times 4\times 2^2=8m$


Displacement,
$r=\sqrt{x^2+y^2}=\sqrt{6^2+8^2}=10m$

$\tan \theta =\frac{8}{6}=\frac{4}{3}\Rightarrow \theta ={53}^{\circ }$

Hence, the displacement is $10m$ at ${53}^{\circ }$ with the positive $x-\text{axis}$.

Final Answer: (10)