Answer to Question #130158 in Physics for Rebecca Omweri

Question

A particle mass 2 units moves in a force field depending on time t given by F=24t^2i+(36t-16)j-12tk.Assuming that at t=0,the particle is located at r=3i+j+4k and a velocity V0=6i+15j-8k.Find the velocity and position at any time t

Answer 1

The acceleration of a particle

$${\bf a}(t)=\frac{{\bf F}(t)}{m}\\ =\frac{(24t^2)\hat i+(36t-16)\hat j-12t\hat k}{2}\\ =(12t^2)\hat i+(18t-8)\hat j-(6t)\hat k$$

The velocity of a particle

$${\bf v}(t)=\int {\bf a}(t)dt+{\bf v}(0)$$ $$=\int\left((12t^2)\hat i+(18t-8)\hat j-(6t)\hat k\right)dt\\+(6\hat i+15\hat j-8\hat k)$$ $$=(4t^3+6)\hat i+(9t^2-8t+15)\hat j-(3t^2+8)\hat k$$

The position of a particle

$${\bf r}(t)=\int {\bf v}(t)dt+{\bf r}(0)$$ $$=\int\left((4t^3+6)\hat i+(9t^2-8t+15)\hat j-(3t^2+8)\hat k\right)dt \\+(3\hat i+\hat j+4\hat k)$$ $$=(t^4+6t+3)\hat i+(3t^3-4t^2+15t+1)\hat j\\ -(t^3+8t-4)\hat k$$