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.
Really, no one?
all you do is take the derivative twice, to get the acceleration function, and then plug in 3.45 s. then use that acceleration to find f in F=ma
To find the net force acting on the particle at t = 3.45 s, we need to determine the particle's acceleration and mass. Then, we can calculate the net force using Newton's second law of motion, which states that the net force (F_net) acting on an object is equal to the product of its mass (m) and acceleration (a): F_net = m * a.
In this case, we are given the position function x(t) = -13.00 + 2.00t + 3.50t^2 - 2.50t^3. To find the acceleration, we need to take the second derivative of this position function with respect to time (t). The first derivative will give us the velocity, and the second derivative will give the acceleration.
Let's start by finding the velocity function, which is the derivative of the position function x(t) with respect to time (t):
v(t) = dx/dt = d/dt (-13.00 + 2.00t + 3.50t^2 - 2.50t^3)
Differentiating each term separately:
v(t) = 0 + 2.00 + 2 * 3.50t - 3 * 2.50t^2
Simplifying:
v(t) = 2.00 + 7.00t - 7.50t^2
Now, let's find the acceleration by taking the derivative of the velocity function v(t) with respect to time (t):
a(t) = dv/dt = d/dt (2.00 + 7.00t - 7.50t^2)
Differentiating each term:
a(t) = 0 + 7.00 - 2 * 7.50t
Simplifying:
a(t) = 7.00 - 15.00t
Now that we have the acceleration function, we can substitute t = 3.45 s to find the acceleration at that time:
a(3.45) = 7.00 - 15.00 * 3.45
Calculating:
a(3.45) = 7.00 - 51.75
a(3.45) ≈ -44.75 m/s^2
We still need the mass (m) of the particle. Unfortunately, it is not provided in the question. Without that information, we cannot calculate the net force on the particle.