You don't need the velocity to answer the question. You need the acceleration at a particular time.
This is not a constant-acceleration problem.
The acceleration is the second derivative of the x(t) function,
a(t) = x''(t) = 7.00 - 15 t
Multiply the value of a at 3.45 s by the mass to get the force acting at that time.
A 0.150 kg particle moves along an x axis according to x(t) = -13.00 + 2.00t + 3.50t2 - 2.50t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.45 s?
I tried solving using s=Vit + (.5)at^2 but I can't find the initial velocity of the particle. All you can find are the times and distances, you can find the average velocity from that but there is an acceleration so that's no good.
3 answers
Thank you, I really appreciate that help. We haven't done derivatives yet and its hard to see without that knowledge...
The first derivative (velocity) is
v(t) = x'(t) = 2 + 7t -7.5 t^2
That tells you that the starting velocity (at t=0) is 2 m/s.
I am surprised they would assign this problem to a class that has not studied derivatives.
v(t) = x'(t) = 2 + 7t -7.5 t^2
That tells you that the starting velocity (at t=0) is 2 m/s.
I am surprised they would assign this problem to a class that has not studied derivatives.