**A 0.32 kg particle moves in an xy plane according to x(t) = - 15 + 2 t - 5 t3 and y(t) = 16 + 7 t - 10 t2, with x and y in meters and t in seconds. At t = 1.2 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and © what is the angle of the particle’s direction of travel?**

**Answer:**

In this first find acceleration using x(t) and y(t)

i.e.,dx2/dt2 =ax=-30*t and dy2/dt2=ay=-20;
Force at t=1.2s
Fx=m*ax=- .32

*30*1.2=-11.52 newton

Fy=m

*ay=-.32*20=-6.4 newton

a.)Fresultant=square root of(Fx+Fy)=magnitude of Fr=13.178 newton;

b)Both fx and fy are negative so 3rd qudrant angle=

-(90+tan_inverse(6.4/11.52))=-119 degrees.

c) velocity can be obtained by differentiating x(t) and y(t) once ;

x(t)=2-10t^2=-12.4 y(t)=7-20*t=-17;

direction =tan_inverse(17/12.4)=58.9 degrees

ans=-(90+58.9)=-143.9 degrees (3rd quadrant)