Question

The path of motion of a 8-lb particle in the horizontal plane is described in terms of polar coordinates as r = (2 t + 1)ft and θ = (0.2 t2 − t) rad, where t is in seconds.

Answer 1

Missing question in the text:
"Determine the magnitude of the resultant force acting on the particle when t=2s."

For Newton's second law, the force is given by
$F=ma$
where m is the mass of the particle and a its acceleration.

The acceleration in polar coordinates is given by:
$a= \sqrt{a_r^2+a_{\theta}^2}$
where
$a_r = r''-r \theta'^2$
$a_{\theta} = r \theta'' +2r'\theta '$
are the radial and tangential accelerations.

Therefore, to find the acceleration starting from the polar coordinates, we need to find the derivatives of r and $\theta$. Let's compute them:
$r=2t+1$
$r'=2$
$r''=0$
We need the values at t=2s, therefore we have
$r=5 ft=1.52 m$
$r'=2 ft=0.61 m$
$r''=0 ft = 0 m$

and 
$\theta = 0.2 t^2-t$
$\theta' = 0.4t-1$
$\theta''=0.4$
And using t=2s,
$\theta = -1.2 rad$
$\theta' = -0.2 rad$
$\theta''=0.4 rad$

And now we can calculate the acceleration:
$a_r=r''-r\theta '^2 =-0.06$
$a_{\theta}=r\theta ''+2r' \theta ' = 0.36$
and
$a= \sqrt{a_r^2+a_{\theta}^2}= \sqrt{(-0.06)^2+(0.36)^2}=0.35 m/s^2$
And using the mass: 
$m=8 lb=3.63 kg$
we find the force:
$F=ma=3.63 kg\cdot 0.35 m/s^2=1.29 N$