Question

due to a force field a particle of mass 5 grams moves along a space curve whose position vector is given as a function of time by
r=(2t^3+t)i + (3t^4-t^2)j -12t^2 k
Find
velocity
Momentum
Acceleration
The force field at any time

Answer 1

$$\bold{v}=(6t^2+1)\bold{i} + (12t^3-2t)\bold{j}-24t \bold{k}$$

$$\bold{p}=0.005(6t^2+1)\bold{i} + 0.005(12t^3-2t)\bold{j}\\-24(0.005)t \bold{k}$$

$$\bold{a}=(12t)\bold{i} + (36t^2-2)\bold{j}-24 \bold{k}$$

$$\bold{F}=0.005(12t)\bold{i} + 0.005(36t^2-2)\bold{j}-24(0.005) \bold{k}$$