The position at time t is
x(t) = 0.5t³ - 3t² + 3t + 2
When the velocity is zero, the derivative of x with respect to t is zero. That is,
x' = 1.5t² - 6t + 3 = 0
or
t² - 4t + 2 = 0
Solve with the quadratic formula.
t = (1/2) [ 4 +/- √(16 - 8)] = 3.4142 or 0.5858 s
When t =0.5858 s, the position is
x = 0.5(0.5858³) - 3(0.5858²) + 3(0.5858) + 2 = 2.828 m
When t=3.4142 s, the position is
x = 0.5(3.4142³) - 3(3.4142²) + 3(3.4142) + 2 = -2.828 m
Reject the negative answer.
Answer:
The velocity is zero when t = 0.586 s, and the distance is 2.83 m
When the acceleration is zero, the second derivative of x with respect to t is zero. That is,
3t - 6 = 0
t = 2
The distance traveled is
x = 0.5(2³) - 3(2²) + 3(2) + 2 = 0
Answer:
When the acceleration is zero, t = 2 s, and the distance traveled is zero.
Sorry but you tell me not to ask for help or answers but you want them yourself
An atom consists of the nucleus (a combination of the proton and neutron) and the electron which orbits the nucleus.
Answer:
7.5 N/m
Explanation:
Potential energy of a spring can be calculated using below formula
Potential energy= 1/2kx^2
potential energy = 60 J
X= displacement = 4 m
K= spring constant=?
Substitute the values we have
60= 1/2 × K × 4^2
60= 1/2 × K × 16
60= K × 8
K= 7.5 N/m
Hence, the spring constant of the spring is 7.5 N/m