Question

A particle starts from origin at$t=0$ when a velocity $5.01m/s$and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3.0\hat{i}+2.0\hat{j})m/s^2$
$\left(a\right)$ what is the $y-$ co-ordinate of the particle at the instant its $x-$ coordinate is $84m?$
$\left(b\right)$ What is the speed of the particle at this time?

Answer 1

${\overrightarrow{U}}_x=5\hat{i}\left(m/s\right)$

${\overrightarrow{U}}_y=0\left(m/s\right)$

${\overrightarrow{a}}_x=3\hat{i}\left(m/s^2\right)$

${\overrightarrow{a}}_y=2\hat{j}\left(m/s^2\right)$

$\left(a\right)$ ${\overrightarrow{S}}_x={\overrightarrow{U}}_{x}t+\frac{1}{2}{\overrightarrow{a}}_x{t}^2$

$84\hat{i}=5t\hat{i}+\frac{1}{2}\times 3\times t^2\hat{i}$

$\therefore 3t^2+10t=168$

$3t^2+10t-168=0$

$3t^2+28t-18t-168=0$

$t(3t+28)-6(3t^2+28)=0$

$(t-6)(3t^2+28)=0$

$\because$ time can't be negative.

$\therefore t=6$

At $t=6,{\overrightarrow{S}}_x=84\hat{i}\left(m\right)$

$\therefore {\overrightarrow{S}}_y=0+\frac{1}{2}{\overrightarrow{a}}_y{t}^2\hat{j}$

$=\frac{1}{2}\times 2\times 36\hat{j}$

${\overrightarrow{S}}_y=36\hat{j}$

$\therefore Y-$ coordinate at $(t-6)\Rightarrow 36m$

$\left(b\right)$ ${\overrightarrow{V}}_x={\overrightarrow{U}}_x+{\overrightarrow{a}}_{x}t$

$5+3\times 6$

$=23i\left(m/s\right)$

${\overrightarrow{V}}_y=0+2\times 6\hat{j}$

$=12\hat{j}\left(m/s\right)$

$|V|=\sqrt{{23}^2+{12}^2}$

$25.94m/s$$

$|V|\approx 26m/s$

Where $|V|=$ Speed of particle at $t=6)$