Starting from the origin at time $t=0$, with initial velocity $5 \hat{j} \mathrm{~m} \mathrm{~s}^{-1

Question

Starting from the origin at time $t=0$, with initial velocity $5 \hat{j} \mathrm{~m} \mathrm{~s}^{-1}$, a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{i}+4 \hat{j}) \mathrm{ms}^{-2}$. At time $t$, its coordinates are $\left(20 \mathrm{~m}, y_0 \mathrm{~m}\right)$. The values of $t$ and $y_0$ are, respectively

Professor PC Thomas Classes · Answer

(d) : Given, initial velocity, $u=5 \hat{j} \mathrm{~m} / \mathrm{s}$ and acceleration, $$ a=(10 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}^2 $$ Also, at $t=t$, position coordinates $(x, y)=\left(20 \mathrm{~m}, y_0 \mathrm{~m}ight)$ Now, from equation of motion, $x=u_x t+\frac{1}{2} a_x t^2$ Here, $u_x=0, x=20 \mathrm{~m}$ and $a_x=10 \mathrm{~m} / \mathrm{s}^2$ $$ 	herefore \quad 20=0+\frac{1}{2} 	imes 10 	imes t^2 	ext { or } t=2 \mathrm{~s} $$ Now, $y=u_y t+\frac{1}{2} a_y t^2$ where $u_y=5 \mathrm{~m} / \mathrm{s}, a_y=4 \mathrm{~m} / \mathrm{s}^2$ and $t=2 \mathrm{~s}$ \ $ 	herefore \quad y=5 	imes 2+\frac{1}{2} 	imes 4 	imes 4=18 \mathrm{~m} $