Question

A particle moves in the \(x-y\) plane under acceleration \(\overrightarrow{a}=3\hat{i}+4\hat{j}~m/s^2\). If the initial velocity is \(\bar{u}=6\hat{i}+8\hat{j}\). Find the magnitude of velocity (in \(m/s\) ) of a particle after \(2~s\)

Answer 1

Given, initial velocity of the particle,
\(\overrightarrow{u}=6\hat{i}+8\hat{j}=u_x\hat{i}+u_y\hat{j}\)
\(u_x=6~m/s,~u_y=8~m/s\)
acceleration, \(\overrightarrow{a}=3~m/s^2,~a_y=4~m/s^2\)
From first equation of motion i.e.
\(v=u+at\)
Along \(x:v_x=u_x+a_xt=6+3\times2=12~m/s\)
Along \(y:v_x=v_y+a_yt=8+4\times2=16~m/s\)
After \(2 ~s\)
Velocity \(\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}=12\hat{i}+16\hat{j}\)
$\left(\underset{V}{\to }\right|=\sqrt{{12}^2+{16}^2}=\sqrt{144+256}=\sqrt{400}=20m/s$
Final Answer: \(20\)