Question

At time t = 0, a particle is at (2m, 4m). It starts moving towards positive x-axis with constant acceleration 2 m/$s^2$ (initial velocity = 0). After 2 s, an additional acceleration of 4 m/$s^2$ starts acting on the particle in negative y-direction also. Find after next 2 s
(a) velocity and (b) coordinate of particle.

Answer 1

$x_o=2cm$

$y_o=4cm$

after $2$ second $x=x_o+\frac{1}{2}a{t}^2=2+\frac{1}{2}\times 2\times 4=6cm$

$y{=}_o=4m$

Now $a=4m{s}^2$ in $-y$ & $a$ is also present

So

$x_1=x+\frac{1}{2}a{r}^2=6+\frac{1}{2}\times 2\times 4=10cm$

$y_1=y_o-\frac{1}{2}a{t}_1^2=4-\frac{1}{2}\times 4\times 4=-4m$

$\Rightarrow$ Coordinates $\equiv (10,-4)$

velocity $v_y=(0+a{t}_1)(-\hat{y})=-\frac{m}{s}\hat{y}$

$v_x=0+a(t+t_1)=0+2\left(4\right)=8\frac{m}{s}\hat{x}$

Net velocity $=\sqrt{V{y}^2+V{x}^2}=\sqrt{64+64}=8\sqrt{2}\frac{m}{s}$

at angle $-{45}^o$ to $+y$ axis