Question
A 0.150 kg particle moves along an x axis according to x(t) = -13.00 + 2.00t + 4.50t2 - 3.00t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.60 s?
Answers
drwls
Compute the second derivative x"(t) at t = 3.60 s. That will be the acceleration at that time.
x'(t) = 2 + 9 t - 9 t^2
x"(t) = 9 - 18 t
x"(t=3.6) = -55.8 m/s^2
Multiply that by the mass for the force. It will be in the -x direction.
x'(t) = 2 + 9 t - 9 t^2
x"(t) = 9 - 18 t
x"(t=3.6) = -55.8 m/s^2
Multiply that by the mass for the force. It will be in the -x direction.