# CHAPTER 2:KINEMATICS OF PARTICLES¶

## SAMPLE PROBLEM 2/1,PAGE NUMBER:27¶

In [1]:
import math
# Variable declaration
# s=2t**3-24t+6;
v_a=72;# Velocity in m/s
v_b=30;# Velocity in m/s
t_0=1;# s
t_1=4;# s

# Calculation
# v=6t**2-24;
# a=12t;
# (a)
t=math.sqrt((v_a+24)/6);# Time in s
# (b)
a=math.sqrt((v_b+24)/6);# Time in s
# (c)
s4=((2*t_1**3)-(24*t)+6);# m
s1=((2*t_0**3)-(24*t_0)+6)# m;
deltaS=s4-s1;# The net displacement during the specified interval in m
print"\n(a)The time required for the particle to reach a velocity of 72 m/s from its initial condition at t=0 is %1.0f s.\n(b)The acceleration of the particle a=%2.0f m/s**2 \n(c)The net displacement,deltaS=%2.0f m"%(t,a,deltaS);

(a)The time required for the particle to reach a velocity of 72 m/s from its initial condition at t=0 is 4 s.
(b)The acceleration of the particle a= 3 m/s**2
(c)The net displacement,deltaS=54 m


## SAMPLE PROBLEM 2/2,PAGE NUMBER:28¶

In [2]:
import math
# Variable declaration
v_x=50;# The initial velocity in ft/sec
a_x=-10;# The acceleration in ft/sec**2
t_0=8;# s
t_1=12;# s

# Calculation
# v_x=90-10t; ft/sec
v_x0=(90-(10*t_0));# The velocity in ft/sec
v_x1=(90-(10*t_1));# The velocity in ft/sec
# x=-5t**2+90t-80; ft
x_0=(-5*t_0**2)+(90*t_0)-80;# ft
x_1=(-5*t_1**2)+(90*t_1)-80;# ft
# The maximum positive x-coordinate is,then, the value of x for t=9 sec which is
t=9;# sec
x_max=(-5*t**2)+(90*t)-80;# ft
print"\nThe velocity of the particle for the conditions of t=8 sec and t=12 sec,v_x=%2.0f ft/sec & v_x=%2.0f ft/sec"%(v_x0,v_x1)
print"\nThe x-coordinate of the particle for the conditions of t=8 sec and t=12 sec, x=%3.0f ft & x=%3.0f ft"%(x_0,x_1)
print"\nThe maximum positive x-coordinate reached by the particle,x_max=%3.0f ft"%x_max

The velocity of the particle for the conditions of t=8 sec and t=12 sec,v_x=10 ft/sec & v_x=-30 ft/sec

The x-coordinate of the particle for the conditions of t=8 sec and t=12 sec, x=320 ft & x=280 ft

The maximum positive x-coordinate reached by the particle,x_max=325 ft


## SAMPLE PROBLEM 2/4,PAGE NUMBER:30¶

In [3]:
%matplotlib inline
import math
import matplotlib.pyplot as plt
from matplotlib.pyplot import plot,suptitle,xlabel,ylabel
import numpy as np
# Variable declaration
t=[0,2*60**-1,4*60**-1,6*60**-1,8*60**-1,10*60**-1];# Time in hrs

# Calculation
# v=8/(1+6t)
v=[8*(1+(6*(t[0])))**-1,8*(1+(6*(t[1])))**-1,8*(1+(6*(t[2])))**-1,8*(1+(6*(t[3])))**-1,8*(1+(6*(t[4])))**-1,8*(1+(6*(t[5])))**-1];# The velocity in knots
t=[0,2,4,6,8,10];# Time in min
plt.figure(1)
plt.subplot(211)
plt.plot(t,v,color='green')
xlabel('t,min')
ylabel('v,knots')
#The distance is obtained by substituting the expression for v into the definition.v=ds/dt and integrating.
t=[0,2*60**-1,4*60**-1,6*60**-1,8*60**-1,10*60**-1];# Time in hrs
# s=4/3 ln(1+6t)
s=[(4*3**-1)*np.log(1+(6*t[0])),(4*3**-1)*np.log(1+(6*t[1])),(4*3**-1)*np.log(1+(6*t[2])),(4*3**-1)*np.log(1+(6*t[3])),(4*3**-1)*np.log(1+(6*t[4])),(4*3**-1)*np.log(1+(6*t[5]))];# The distance s in nautical miles
t=[0,2,4,6,8,10];# Time in min
plt.subplot(212)
plt.plot(t,s,color='red')
xlabel('t,min')
ylabel('s,mi(nautical)')
plt.show()


## SAMPLE PROBLEM 2/5,PAGE NUMBER:46¶

In [4]:
%matplotlib inline
import math
from matplotlib.pyplot import plot,suptitle,xlabel,ylabel
import numpy as np
# Variable declaration
# v_x=50-60t;
# y=100-4t**2;
# where v_x is in meters per second, y is in meters, and t is in seconds.

# Calculation
# x=50t-8t**2;
a_x=-16;# The x-component of the acceleration in m/s**2
# v_y=-8t; The y-component of the velocity in m/s
a_y=-8;# The y-component of the acceleration in m/s**2
# When y=0,
t=math.sqrt(100/4);
v_x=50-(16*t);
v_y=-8*(t);
v=math.sqrt((v_x**2)+(v_y**2));# m/s
a=math.sqrt(a_x**2+a_y**2);# m/s**2
print"\nThe velocity,v=%2.0fi+(%2.0fj) m/s \nThe acceleration,a=%2.0fi+(%1.0fj) m/s**2"%(v_x,v_y,a_x,a_y);
y=[0,20,40,60,80,100];# m
t=[math.sqrt((100-y[0])/4),math.sqrt((100-y[1])/4),math.sqrt((100-y[2])/4),math.sqrt((100-y[3])/4),math.sqrt((100-y[4])/4),math.sqrt((100-y[5])/4)];# s
x=[((50*t[0])-(8*t[0]**2)),((50*t[1])-(8*t[1]**2)),((50*t[2])-(8*t[2]**2)),((50*t[3])-(8*t[3]**2)),((50*t[4])-(8*t[4]**2)),((50*t[5])-(8*t[5]**2))];# m
v_x=[50-(16*(t[0])),50-(16*t[1]),50-(16*t[2]),50-(16*t[3]),50-(16*t[4]),50-(16*t[5])];
v_y=[-8*(t[0]),-8*(t[1]),-8*(t[2]),-8*(t[3]),-8*(t[4]),-8*(t[5])];
v=[math.sqrt((v_x[0]**2)+(v_y[0]**2)),math.sqrt((v_x[1]**2)+(v_y[1]**2)),math.sqrt((v_x[2]**2)+(v_y[2]**2)),math.sqrt((v_x[3]**2)+(v_y[3]**2)),math.sqrt((v_x[4]**2)+(v_y[4]**2)),math.sqrt((v_x[5]**2)+(v_y[5]**2))],;# m/s
a=math.sqrt(a_x**2+a_y**2);# m/s**2
plot(x,y,'*-',color='green')
xlabel('x,m')
ylabel('y,m')


The velocity,v=-30i+(-40j) m/s
The acceleration,a=-16i+(-8j) m/s**2

Out[4]:
<matplotlib.text.Text at 0x6c97198>

## SAMPLE PROBLEM 2/6,PAGE NUMBER:47¶

In [5]:
import math
from scipy.optimize import fsolve
# Variable declaration
v_0=80;# The launch speed in ft/sec
theta=35;# The launch angle in degree
m=8;# lb
g=32.2;# The acceleration due to gravity in ft/sec**2
y_0=6;# ft
x_0=0;# ft
x=100+30;# ft

# Calculation
v_x0=v_0*math.cos(theta);# ft/sec
t=(x-x_0)/v_x0;# s
v_y0=v_0*math.sin(theta);# ft/sec
y=(y_0+(v_y0*t))-((1/2)*g*t**2);# ft
# (a)
# We now find the flight time by setting
y_01=20;# ft
def equations(p):
t_f=p
return((y_0+(v_y0*t_f)-((1/2)*g*t_f**2)))-y_01;
t_f=fsolve(equations,(1))
x=x_0+(v_x0*t_f);# ft
print"\n(a)The time duration of the flight,t_f=%1.2f s"%t_f;
#(b)
print"\n(b)Thus the point of first impact is (x,y)=(%3.0f,%2.0f)ft"%(x,y_01);
# (c)
v_y=0;# ft
h=((v_y0**2-v_y**2)/(2*g))+6;# ft
print"\n(c)The maximum height above the horizontal field attained by the ball,h=%2.1f ft"%h;
# (d)
v_x=v_x0;# ft/sec
v_y=v_y0-(g*t_f);# ft/sec
print"\n(d)The impact velocity,v=%2.1f i+(%2.1f j) ft/sec"%(v_x,v_y);
x=100+30;# ft (given)
v_0=75;# ft/sec (given)
v_x0=v_0*math.cos(theta);# ft/sec
t=(x-x_0)/v_x0;# s
v_y0=v_0*math.sin(theta);# ft/sec
y=(y_0+(v_y0*t))-((1/2)*g*t**2);# ft
print"\n   The point of impact is (x,y)=(%3.0f,%2.1f)ft"%(x,y);

(a)The time duration of the flight,t_f=-0.41 s

(b)Thus the point of first impact is (x,y)=( 30,20)ft

(c)The maximum height above the horizontal field attained by the ball,h=24.2 ft

(d)The impact velocity,v=-72.3 i+(-21.1 j) ft/sec

The point of impact is (x,y)=(130,67.6)ft


## SAMPLE PROBLEM 2/7,PAGE NUMBER:57¶

In [6]:
import math
# Variable declaration
a=3;# m/s**2
v_A=100;# km/h
v_C=50;# km/h
s=120;# m

# Calculation
v_A=((v_A*(1000)*3600**-1));# The velocity in m/s
v_C=((v_C*(1000)*3600**-1));# The velocity in m/s
a_t=((v_C**2-v_A**2)/(2*s));# The acceleration in m/s**2
# (a) Condition at A.
a_n=math.sqrt(a**2-a_t**2);# The acceleration in m/s**2
rho_A=v_A**2*a_n**-1;# The radius of curvature at A in m
# (b) Condition at B.
a_n=0;# m/s**2
a_b=a_n+a_t;# The acceleration at the inflection point B in m/s**2
# (c) Condition at C.
rho=150;# The radius of curvature of the hump at C in m
a_n=v_C**2/rho;# The normal acceleration in m/s**2
a=math.sqrt(a_n**2+a_t**2);# The total acceleration at C in m/s**2
print"\n(a)The radius of curvature at A,rho=%3.0f m"%rho_A
print"\n(b)The acceleration at the inflection point B,a=%1.2f m/s**2"%a_b
print"\n(c)The total acceleration at C,a=%1.2f m/s**2"%a

(a)The radius of curvature at A,rho=432 m

(b)The acceleration at the inflection point B,a=-2.41 m/s**2

(c)The total acceleration at C,a=2.73 m/s**2


## SAMPLE PROBLEM 2/8,PAGE NUMBER:58¶

In [7]:
import math
# Variable declaration
g=30;# The acceleration due to gravity in ft/sec**2
theta=15;# The direction of its trajectory in degree
v=12000;# The velocity in mi/hr
a_x=20;# The horizontal component of acceleration in ft/sec**2
a_y=g;# The downward acceleration component in ft/sec**2

# Calculation
a_n=(a_y*math.cos(theta*math.pi/180))-(a_x*math.sin(theta*math.pi/180));# The normal component of acceleration in ft/sec**2
a_t=(a_y*math.sin(theta*math.pi/180))+(a_x*math.cos(theta*math.pi/180));# The tangential component of acceleration in ft/sec**2
# (a)
v=v*44/30;# ft/sec
rho=v**2/a_n;# The radius of curvature in ft
# (b)
vdot=a_t;# The t-component of acceleration in ft/sec**2
# (c)
# (d)
a=[a_n,a_t];# The total acceleration in ft/sec**2
print"\n(a)The radius of curvature,rho=%2.2e ft \n(b)The t-component of acceleration,v_dot=%2.1f ft/sec**2 \n(c)The angular rate of line GC,betadot=%2.2e rad/sec \n(d)The total acceleration,a=%2.1f e_n+%2.1f e_t ft/sec**2"%(rho,vdot,betadot,a[0],a[1]);

(a)The radius of curvature,rho=1.30e+07 ft
(b)The t-component of acceleration,v_dot=27.1 ft/sec**2
(d)The total acceleration,a=23.8 e_n+27.1 e_t ft/sec**2


## SAMPLE PROBLEM 2/9,PAGE NUMBER:69¶

In [8]:
import math
# Variable declaration
# theta=0.2t+0.02t**3;
# r=0.2+0.04t**2;
t=3;# s

import math
# Variable declaration
# theta=0.2t+0.02t**3;
# Variable declaration
# theta=0.2t+0.02t**3;
# r=0.2+0.04t**2;
t=3;# s

# Calculation
r_3=0.2+(0.04*t**2);# m
rdot_3=0.08*t;# m/s
rdotdot_3=0.08;# m/s**2
v_r=rdot_3;# m/s
v=math.sqrt(v_r**2+v_theta**2);# m/s
a=math.sqrt(a_r**2+a_theta**2);# m/s**2
print"\nThe magnitudes of the velocity and acceleration of the slider, v=%0.3f m/s and a=%0.3f m/s**2"%(v,a);

The magnitudes of the velocity and acceleration of the slider, v=0.479 m/s and a=0.601 m/s**2


## SAMPLE PROBLEM 2/10,PAGE NUMBER:70¶

In [9]:
# Variable declaration
theta_i=30;# degrees
r=25*10**4;# ft
rdot=4000;# ft/sec
theta=0.80;# deg/sec
g=31.4;# ft/sec**2

# Calculation
v_r=rdot;# ft/sec
v_theta=r*(theta*math.pi/180);# ft/sec
v=math.sqrt(v_r**2+v_theta**2);# ft/sec
a_r=-g*math.cos(theta_i*math.pi/180);# ft/sec**2
a_theta=g*math.sin(theta_i*math.pi/180);# ft/sec**2
rdotdot=a_r+(r*(theta*(math.pi/180))**2);# ft/sec**2

The velocity of the rocket,v=5309 ft/sec


## SAMPLE PROBLEM 2/12,PAGE NUMBER:83¶

In [10]:
import math
# Variable declaration
v_0=250;# km/h
theta_i=15;# degree
a=0.8;# m/s**2
t=60;# seconds
s_0=0;# m
x=3000;# m

# Calculation
# (a)
v_0=v_0/3.6;# m/s
v=v_0+(a*t);# m/s
s=s_0+(v_0*t)+((a*t**2)/2);# m
y=s*math.cos(theta_i*math.pi/180);# m
theta=math.atan(y/x)*180/math.pi;# degree
r=math.sqrt(x**2+y**2);# m
v_xy=v*math.cos(theta_i*math.pi/180);# m/s
v_r=v_xy*math.sin(theta*math.pi/180);# m/s
v_theta=v_xy*math.cos(theta*math.pi/180);# m/s
zdot=v*math.sin(theta_i*math.pi/180);# m/s
v_z=zdot;# m/s
# (b)
z=y*math.tan(theta_i*math.pi/180);# m
phi=math.degrees(math.atan(z/r));# degree
R=math.sqrt(r**2+z**2);# m
v_R=(v_r*math.cos(phi*math.pi/180))+(zdot*math.sin(phi*math.pi/180));# m/s
v_phi=(zdot*(math.cos(phi*math.pi/180)))-(v_r*math.sin(phi*math.pi/180));# m/s
phidot=v_phi/R;# m/s

(a)v_r=99.2 m/s
zdot=v_z=30.4 m/s
(b)v_R=103.6 m/s


## SAMPLE PROBLEM 2/13,PAGE NUMBER:90¶

In [11]:
import math
from scipy.optimize import fsolve
# Variable declaration
v_A=800;# km/h
theta_1=45*math.pi/180;# degree
theta_2=60*math.pi/180;# degree
theta_3=75*math.pi/180;# degree

# Calculation
# [i] Graphical.
v_BA=586;# km/h
v_B=717;# km/h
print"nv_BA=%3.0f km/h and v_B=%3.0f km/h"%(v_BA,v_B);
# (II) Trigonometric.
v_B=(math.sin(theta_2)*v_A)/math.sin(theta_3);# km/h
print"\nv_B=%3.0f km/h"%v_B;
# (III) Vector Algebra
def equations(p):
v_BA,v_B=p
return((v_A-(v_BA*math.cos(theta_2)))-(v_B*math.cos(theta_1)),(v_B*math.sin(theta_1))-(v_BA*math.sin(theta_2)));
v_BA,v_B=fsolve(equations,(1,1))
print"\nv_BA=%3.0f km/h and v_B=%3.0f km/h"%(v_BA,v_B);

nv_BA=586 km/h and v_B=717 km/h

v_B=717 km/h

v_BA=586 km/h and v_B=717 km/h


## SAMPLE PROBLEM 2/14,PAGE NUMBER:91¶

In [12]:
import math
# Variable declaration
v_A=45;# mi/hr
v_B=30;# mi/hr
a_A=3;# ft/sec**2
theta_1=30;# degree
theta_2=60;# degree
rho=440;# The radius of curvature in ft

# Calculation
# Velocity
v_A=v_A*(5280*3600**-1);# ft/sec
v_B=v_B*(5280*3600**-1);# ft/sec
# By the application of the law of cosines and the law of sines gives
v_BA=math.sqrt(v_A**2+v_B**2-(2*v_A*v_B*math.cos(theta_2*math.pi/180)));# ft/sec
theta=math.degrees(math.asin((v_B*math.sin(theta_2*math.pi/180))/v_BA));# degree
# Acceleration
a_B=(v_B)**2/rho;# ft/sec**2
a_BAx=a_B*math.cos(theta_1*math.pi/180)-a_A;# ft/sec**2
a_BAy=a_B*math.sin(theta_1*math.pi/180);# ft/sec**2
a_BA=math.sqrt(a_BAx**2+a_BAy**2);# ft/sec**2
beta=180-math.degrees((math.asin((a_B/a_BA)*math.sin(theta_1*math.pi/180))));# degree
print"\nv_BA=%2.1f ft/sec \ntheta=%2.1f degree \na_AB=%1.2f ft/sec**2 \nbeta=%2.1f degree"%(v_BA,theta,a_BA,beta);

v_BA=58.2 ft/sec
theta=40.9 degree
a_AB=2.34 ft/sec**2
beta=110.2 degree


## SAMPLE PROBLEM 2/15,PAGE NUMBER:100¶

In [13]:
import math
# Variable declaration
v_A=0.3;# m/s

# Calculation
# Solution [i].
# v_A=y_A,v_B=y_B
v_B=-(2*v_A)/3;# m/s
print"\nThe velocity of B,v_B=%0.1f m/s"%v_B;
# Solution (II).
v_B=abs((2*v_A)/3);# m/s
print"\nThe velocity of B,v_B=%0.1f m/s (upward)"%v_B;

The velocity of B,v_B=-0.2 m/s

The velocity of B,v_B=0.2 m/s (upward)