
import math
import numpy as np

def func(x, A=10, a=1, b=1, c=1, d=1):
    return A * np.log(c + a*x) - (d + b*x)


def bisect(A,a,b,c,d,tol=1e-6,x1=1,x2=2):

    y = tol
    it = 0
    
    while abs(y) >= tol:
        y1 = func(x1,A=A, a=a, b=b, c=c, d=d)
        y2 = func(x2,A=A, a=a, b=b, c=c, d=d)
        x = (x1 + x2) / 2
        y = func(x,A=A, a=a, b=b, c=c, d=d)
        
        if np.sign(y) == np.sign(y1):
            x1 = x
        elif np.sign(y) == np.sign(y2):
            x2 = x
            
        it+=1
        
        print("Solution at iteration {} = {}".format(it, x))
        
    return x

print(bisect(196,-.001,-.001,30.02,30.02,tol=1e-6,x1=0,x2=30000))

Solution at iteration 1 = 15000.0
Solution at iteration 2 = 22500.0
Solution at iteration 3 = 26250.0
Solution at iteration 4 = 28125.0
Solution at iteration 5 = 29062.5
Solution at iteration 6 = 28593.75
Solution at iteration 7 = 28828.125
Solution at iteration 8 = 28945.3125
Solution at iteration 9 = 29003.90625
Solution at iteration 10 = 29033.203125
Solution at iteration 11 = 29018.5546875
Solution at iteration 12 = 29011.23046875
Solution at iteration 13 = 29014.892578125
Solution at iteration 14 = 29013.0615234375
Solution at iteration 15 = 29013.97705078125
Solution at iteration 16 = 29014.434814453125
Solution at iteration 17 = 29014.663696289062
Solution at iteration 18 = 29014.77813720703
Solution at iteration 19 = 29014.835357666016
Solution at iteration 20 = 29014.863967895508
Solution at iteration 21 = 29014.84966278076
Solution at iteration 22 = 29014.856815338135
Solution at iteration 23 = 29014.86039161682
Solution at iteration 24 = 29014.858603477478
Solution at iteration 25 = 29014.857709407806
Solution at iteration 26 = 29014.858156442642
Solution at iteration 27 = 29014.85837996006
Solution at iteration 28 = 29014.85849171877
Solution at iteration 29 = 29014.858547598124
Solution at iteration 30 = 29014.8585755378
Solution at iteration 31 = 29014.858561567962
Solution at iteration 32 = 29014.858554583043
29014.858554583043


import numpy as np
import pylab as pl

Mb = 20 # Mb equals the mass of the empty bucket in grams.

M0 = 20 * 1000 # M0 equals the mass of the bucket at time zero in kilograms.

m = 10 * 1000 # Mass of the weight in kilograms

g = 9.8 # Acceleration of gravity
At = 0 # At is the starting point for Acceleration at time t.
tbeg = 0 # Change this value to the beginning time frame for the Leakage.
tend = 30000 # Change this value to the ending time frame for the Leakage.
t = tbeg # Setting the beginning time
L = 1 # Lambda equals rate of Leakage at 1 gram/s
Lt = 0 # Lambda equals total Leakage at time t
# t1 = np.linspace(0.0, 5.0)

pl.rcParams.update({'font.size': 18}) # Sets fontsize to 18
pl.figure(figsize=(12,10)) # Sets figure size to be 12 x 10
pl.xlabel('time (s)') # Label the x-axis as time
pl.ylabel('Acceleration (m/s^2)') # Label the y-axis as Acceleration
pl.title('Acceleration vs time graph')

while t >= tbeg and t <= tend:
    At = ((Mb + M0 - Lt - m)*g) / (Mb + M0 - Lt + m)
    
    pl.xlim(tbeg-1, tend+1) # Expands the graph window size of the x- axis to show the data values clearly.
    
    # Plots t along x axis, and At along y-axis. Uses blue (b) circular (o) markers joined by line
    pl.plot(t, At, 'bo-') 
    
    t += 1
    Lt += 1


import numpy as np
import pylab as pl

Mb = 20 # Mb equals the mass of the empty bucket in grams.

M0 = 20 * 1000 # M0 equals the mass of the bucket at time zero in kilograms.

m = 10 * 1000 # Mass of the weight in kilograms

Vt = 0 # Vt is the starting point for Velocity at time t.
tbeg = 0 # Change this value to the beginning time frame for the Leakage.
tend = 30000 # Change this value to the ending time frame for the Leakage.
t = tbeg # Setting the beginning time
L = 1 # Lambda equals rate of Leakage at 1 gram/s
Lt = 0 # Lambda equals total Leakage at time t
# t1 = np.linspace(0.0, 5.0)
g = 9.8 # Acceleration of gravity
gL = g * L


pl.rcParams.update({'font.size': 18}) # Sets fontsize to 18
pl.figure(figsize=(12,10)) # Sets figure size to be 12 x 10
pl.xlabel('time (s)') # Label the x-axis as time
pl.ylabel('Velocity (m/s)') # Label the y-axis as Velocity
pl.title('Velocity vs time graph')

while t >= tbeg and t <= tend:
    Vt = 2 * gL * m * (math.log(Mb + M0 + m - Lt)) - (gL * (Mb + M0 + m - Lt))
    
    pl.xlim(tbeg-1, tend+1) # Expands the graph window size of the x- axis to show the data values clearly.
    
    # Plots t along x axis, and At along y-axis. Uses blue (b) circular (o) markers joined by line
    pl.plot(t, Vt, 'bo-') 
        
    t += 1
    Lt += 1

University Physics I Project - Mechanics problem¶

By Michael Camilo¶

Intro/Theory¶

Introduction:¶

Free Body Diagram:¶

Acceleration of the system:¶

Velocity of the system/ Max Velocity:¶

Transcendental Functions:¶

Problem and Methods Used:¶

Conclusion:¶