A 0.25 kg particle moves along an x axis according to x(t) = - 10 + 3 t + 2 t2 - 5 t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.6 s ? Give an expression for the (a) x, (b) y and (c) z components.

Question

A 0.25 kg particle moves along an x axis according to x(t) = - 10 + 3 t + 2 t2 - 5 t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.6 s ? Give an expression for the (a)  x, (b)  y and (c)  z components.

bobpursley · Answer

force=mass*acceleration = mass*d"(x)/dt" = mass*d/dt (3+4t-15t^2) = mass* (4-30t) so put in t=3.6, and solve for force. Of course, it is in the x direction. the y component and z component is zero.