Physics problem, please help?
A 0.36 kg particle moves in an xy plane according to x(t) = - 11 + 1 t - 3 t^3 and y(t) = 17 + 4 t - 9 t^2, with x and y in meters and t in seconds. At t = 0.7 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 (c) what is the angle of the particle's direction of travel?
Comments
If the position vector is
r(t) = ( -11 + t - 3t^3 , 17 + 4 t - 9 t^2 )
then the velocity vector follows from differentiation
v( t) = ( 1 - 9 t^2 , 4 - 18 t)
and the acceleration vector follows from one more differentiation
a(t) = ( -18 t , -18) = -18 ( t ,1)
The force vector, from F= m a, so
F = - 0.36 *18 (t,1) = - 6.48 ( t, 1)
(a) with t= 0.7 we find
F= -6.48 (0.7,1) = (-4.54, -6.48)
The magnitude of F therefore is
|F| = sqrt(4.54^2 + 6.48^2) = 7.91 N
( b) tan( angle with negative x- axis) = 6.48/4.54
=> angle with pos x- axes= - 125 degrees
(c) calculate the velocity vector v(0.7) using the expression derived above and calculat the tangent as under b.