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?
Compute the acceleration, x"(t). It equals 7 - 15t m/s^2.
Compute its value at t = 3.34 s
Divide that by the mass for the acceleration.
To find the net force acting on the particle at t = 3.45 s, we need to determine the acceleration of the particle at that time and then multiply it by the mass of the particle.
First, let's find the expression for the acceleration a(t) by taking the second derivative of the position function x(t):
x(t) = -13.00 + 2.00t + 3.50t^2 - 2.50t^3
Taking the derivative with respect to time, we get:
v(t) = dx(t)/dt = 2.00 + 7.00t - 7.50t^2
Taking the derivative of velocity v(t) with respect to time, we get:
a(t) = dv(t)/dt = 7.00 - 15.00t
Now,let's find the acceleration at t = 3.45 s by substituting this value into the expression for acceleration:
a(3.45) = 7.00 - 15.00(3.45)
a(3.45) = 7.00 - 51.75
a(3.45) = -44.75 m/s^2
Now, multiply the acceleration by the mass of the particle:
m = 0.150 kg
F(net) = m * a(3.45)
F(net) = (0.150 kg) * (-44.75 m/s^2)
F(net) = -6.7125 N
So, in unit-vector notation, the net force acting on the particle at t = 3.45 s is approximately -6.7125 N in the x direction.