Question

The position of a 2kg mass is given by x=(2t^3-3t^2)m, where t is in seconds. What is the net horizontal force on the mass at (a)t=0s and (b)t=1s?

Answer 1

x=(2t^3-3t^2

v=dx/dt
v=6t^2-6t

a=dv/dt
a= 12t-6

F=m*a
F=2*(12t-6)
F=24t-12

t=0 => F=-12
t=1 => F=12