a) ![F=(3675i-4543k)N](https://tex.z-dn.net/?f=F%3D%283675i-4543k%29N)
b) 5843 N
Explanation:
a)
The position of the UFO at time t is given by the vector:
![r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k](https://tex.z-dn.net/?f=r%28t%29%3D%280.24t%5E3%2B25%29i%2B%284.2t%29j%2B%28-0.43t%5E3%2B0.8t%5E2%29k)
Therefore it has 3 components:
![r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2](https://tex.z-dn.net/?f=r_x%3D0.24t%5E3%2B25%5C%5Cr_y%3D4.2t%5C%5Cr_z%3D-0.43t%5E3%2B0.8t%5E2)
We start by finding the velocity of the UFO, which is given by the derivative of the position:
![v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t](https://tex.z-dn.net/?f=v_x%3Dr%27_x%3D%5Cfrac%7Bd%7D%7Bdt%7D%280.24t%5E3%2B25%29%3D3%5Ccdot%200.24t%5E2%3D0.72t%5E2%5C%5Cv_y%3Dr%27_y%3D%5Cfrac%7Bd%7D%7Bdt%7D%284.2t%29%3D4.2%5C%5Cv_x%3Dr%27_z%3D%5Cfrac%7Bd%7D%7Bdt%7D%28-0.43t%5E3%2B0.8t%5E2%29%3D-1.29t%5E2%2B1.6t)
And then, by differentiating again, we find the acceleration:
![a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6](https://tex.z-dn.net/?f=a_x%3Dv%27_x%3D%5Cfrac%7Bd%7D%7Bdt%7D%280.72t%5E2%29%3D1.44t%5C%5Ca_y%3Dv%27_y%3D%5Cfrac%7Bd%7D%7Bdt%7D%284.2%29%3D0%5C%5Ca_z%3Dv%27_z%3D%5Cfrac%7Bd%7D%7Bdt%7D%28-1.29t%5E2%2B1.6t%29%3D-2.58t%2B1.6)
The weight of the UFO is W = 12,500 N, so its mass is:
![m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7BW%7D%7Bg%7D%3D%5Cfrac%7B12500%7D%7B9.8%7D%3D1276%20kg)
Therefore, the components of the force on the UFO are given by Newton's second law:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
So, Substituting t = 2 s, we find:
![F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N](https://tex.z-dn.net/?f=F_x%3Dma_x%3D%281276%29%281.44t%29%3D%281276%29%281.44%29%282%29%3D3675%20N%5C%5CF_y%3Dma_y%3D0%5C%5CF_z%3Dma_z%3D%281276%29%28-2.58t%2B1.6%29%3D%281276%29%28-2.58%282%29%2B1.6%29%3D-4543%20N)
So the net force on the UFO at t = 2 s is
![F=(3675i-4543k)N](https://tex.z-dn.net/?f=F%3D%283675i-4543k%29N)
b)
The magnitude of a 3-dimensional vector is given by
![|v|=\sqrt{v_x^2+v_y^2+v_z^2}](https://tex.z-dn.net/?f=%7Cv%7C%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%2Bv_z%5E2%7D)
where
are the three components of the vector
In this problem, the three components of the net force are:
![F_x=3675 N\\F_y=0\\F_z=-4543 N](https://tex.z-dn.net/?f=F_x%3D3675%20N%5C%5CF_y%3D0%5C%5CF_z%3D-4543%20N)
Therefore, substituting into the equation, we find the magnitude of the net force:
![|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N](https://tex.z-dn.net/?f=%7CF%7C%3D%5Csqrt%7B3675%5E2%2B0%5E2%2B%28-4543%29%5E2%7D%3D5843%20N)