Answer:
1) we know
[tex]F=ma\\a=F/m\\a=20/4=3m/s^2[/tex]
since the force is acting upward and the acceleration is in positive z-direction
[tex]a(t)=5k[/tex]
we integrate it to get the velocity
[tex]v(t)=5tk+C[/tex]
where C is integration constant
[tex]Given\\v(0)=i-j[/tex]
using this to find C
[tex]v(0)=5(0)k+C\\i-j=0+C\\C=i-j[/tex]
substitute back the value of C we get:
[tex]v(t)=i-j+5tk[/tex]
integrate to get the position
[tex]r(t)=ti-tj+\frac{5t^2}{2} k+C[/tex]
where C is zero because it starts from origin
speed is magnitude of velocity
[tex]v(t)=i-j+5tk\\therefore\\\\v(t)=\sqrt{1^2+(-1)^2+(5t)^2} =\sqrt{2+25t^2}[/tex]
