Question

A dense particle with mass 5 kg follows the path r(t)=⟨sin(3t),cos(4t),2t9/2⟩r(t)=⟨sin⁡(3t),cos⁡(4t),2t9/2⟩ with units in meters and seconds. What force acts on the mass at t=0t=0?

Answer 1

Answer:

Therefore the force acts on the the mass is $<0,80,\frac{315}{2}>$.

Explanation:

Force: Force is the product of mass and acceleration.

F= ma

F is denoted force

m  is denoted mass

a  is denoted acceleration.

The first order derivative of the path gives the velocity of the particle.

The second order derivative of the path gives the acceleration of the particle.

Rule of derivatives:

1. $\frac{d}{dx}sin\ mx= mcos \ mx$
2. $\frac{d}{dx}cos\ mx= -msin \ mx$
3. $\frac{d}{dx}(ax^n)= a\frac{d}{dx}(x^n)= a n x^{n-1}$

The path is

r(t)= <sin(3t), cos(4t),$2t^{\frac92}$ >

Differentiating with respect t

$v=r'(t)=<3cos \ 3t, -4sin \ 4t, 2.\frac92 t^{\frac92-1}>$

             $=<3cos \ 3t, -4sin \ 4t, 9t^{\frac72}>$

Again differentiating with respect t

$a= r''(t)=<9sin \ 3t, 16cos \ 4t, 9.{\frac72}t^{\frac72-1}>$

              $=<9sin \ 3t, 16cos \ 4t, \frac{63}{2}t^{\frac52}>$

When t=0

$a=r''(0)=<9sin \ (3.0), 16cos \ (4.0), \frac{63}{2}0^{\frac52}>$

              $=<0, 16, \frac{63}{2}}>$

The mass of the object is 5 kg.

Force =$5.<0, 16, \frac{63}{2}}>$

          $=<0,80,\frac{315}{2}>$

Therefore the force acts on the the mass is $<0,80,\frac{315}{2}>$.