Answer
$r(t)= ti-tj+\dfrac{5}{2}t^2k$ and $|v(t)|=\sqrt{25t^2+2}$
Work Step by Step
Given: $m= 4 kg$ and $F(t)=20 N$
In order to calculate the speed of a particle, whose mass is $m$ we will take the help of Newton's Second law of motion:
Force vector: $F(t)=ma(t) \implies a(t)=\frac{F(t)}{m}=\frac{20}{4}=5$ ms^{-2}
Since, $v(t)=\int a(t) dt$ which has direction towards $+z$-axis.
Thus, $v(t)=5t k+A$
Here, A shows a constant of integration.
From question, we have $v(0)=i-j \implies A=i-j$
$v(t)=5t k+i-j$ and $|v(t)|=\sqrt{25t^2+2}$
Since, $r(t)=\int v(t) \implies ti-tj+5(\dfrac{t^2}{2})k$
Hence, the required answer is :
$r(t)= ti-tj+\dfrac{5}{2}t^2k$ and $|v(t)|=\sqrt{25t^2+2}$
