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.
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.
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.
To find the net force acting on the particle, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the acceleration can be obtained by taking the second derivative of the position function x(t).
First, let's find the velocity of the particle. The velocity v(t) is the derivative of the position function x(t) with respect to time:
v(t) = dx(t)/dt = d(-13.00 + 2.00t + 3.50t^2 - 2.50t^3)/dt
= 2.00 + 7.00t - 7.50t^2
To find the initial velocity, we need to evaluate the velocity function v(t) at the initial time t = 0:
v(0) = 2.00 + 7.00(0) - 7.50(0)^2 = 2.00
Next, let's find the acceleration a(t). The acceleration a(t) is the derivative of the velocity function v(t) with respect to time:
a(t) = dv(t)/dt = d(2.00 + 7.00t - 7.50t^2)/dt
= 7.00 - 15.00t
Now, we can find the net force F(t) at any given time t by multiplying the mass of the particle (0.150 kg) with the acceleration a(t):
F(t) = m * a(t)
= 0.150 kg * (7.00 - 15.00t)
To find the net force at t = 3.45 s, substitute t = 3.45 into the expression for F(t):
F(3.45) = 0.150 kg * (7.00 - 15.00 * 3.45)
= 0.150 kg * (-36.75)
Calculating the result:
F(3.45) = -5.5125 N
Therefore, the net force acting on the particle at t = 3.45 s is approximately -5.5125 N in the negative x-direction.
Note: Remember to include the correct units (in this case, meters, seconds, kilograms, and newtons) throughout the calculations.