Answer:
a).[tex]p(x,y,z)=(0.90m,0,0.135m)[/tex]
b).[tex]v(x,y,z)=(3x10^5 a_{x},0,-9x10^4 a_{z})[/tex]
c).[tex]E_{k}=1.5x10^{-5}J[/tex]
Explanation:
The Newton second law using to determinate position
a).
[tex]F=m*a[/tex]
[tex]F=m*a=m*\frac{d^2Z}{dt^2}[/tex]
[tex]\frac{dZ}{dt}=v_{z}=\frac{q*E}{m}*t[/tex]
[tex]z=\frac{q*E}{2*m}*t^2[/tex]
[tex]z=\frac{(0.3x10{-6}*30}{2*(3x10^{-16})}[/tex]
[tex]z=-1.5x10^{10}*t^2m[/tex]
So at t=3us
[tex]z=-(1.5x10^{10})*(3x10^{-6})^2=-0.135m[/tex]
To x at t=3us
[tex]x=v*t=3x10^5*(3x10^{-6})^2=0.90m[/tex]
[tex]p(x,y,z)=(0.90m,0,0.135m)[/tex]
b).
Velocity now, derive the position can get the velocity of the function so
[tex]z=\frac{q*E}{2*m}*t^2[/tex]
[tex]\frac{dz}{dt}=\frac{q*E}{m}*t[/tex]
[tex]v=\frac{q*E}{m}*t[/tex]
at t=3us
[tex]v=\frac{-0.3x10^{-6}*30}{3x10^{-16}}*3x10^{-6}=-9x10^4\frac{m}{s}[/tex]
so
[tex]v=(3x10^5 a_{x},0,-9x10^4 a_{z})[/tex]
c).
Kinetic energy
[tex]E_{k}=\frac{1}{2}*m*v^2[/tex]
[tex]E_{k}=\frac{1}{2}*3x10^{-16}*(1.13x10^5)^2=1.5x10^{-5}J[/tex]
