Answer
$r(t)= ti-tj+\dfrac{5}{2}t^2k$ and $|v(t)|=\sqrt{25t^2+2}$
Work Step by Step
As we are given that $m= 4 kg$ ;$F(t)=20 N$
Use Newton's Second law of motion.
$F(t)=ma(t)$
This yields $a(t)=\frac{F(t)}{m}$
$=\frac{20}{4}$
$=5 ms^{-2}$
As we are given that $v(t)=\int a(t) dt$ directed towards positive $+z$-axis.
Thus, $v(t)=5t k+A$
From the given question: $v(0)=i-j \implies A=i-j$
$v(t)=5t k+i-j$
This implies $|v(t)|=\sqrt{25t^2+2}$
Also, $r(t)=\int v(t) \implies ti-tj+5(\dfrac{t^2}{2})k$
