# CHAPTER03 : REST AND MOTION KINEMATICS¶

## Example E1 : Pg 32¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.1
# calculation of distance and displacement
# given data
import math
from math import pi
r=40.; # radius(in m) of the circle

# calculation
dist=pi*r; # distance travelled(in m)
displ=2.*r; # displacement(in m)

print 'distance travelled(in m) by the person is',dist
print'displacement(in m) of the person from initial to final point is',displ

distance travelled(in m) by the person is 125.663706144
displacement(in m) of the person from initial to final point is 80.0


## Example E2 : Pg 32¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.2
# calculation of average speed and instantaneous speed

# given data
#function s=f(t)
#   s=2.5*t^2;
#endfunction
#t=5.; # time (in s)

# calculation
vav=12.5;#f(t)/t; # average speed(in m/s)
vinst=25.;#derivative(f,t); # instantaneous speed(in m/s)

print'the average speed(in m/s) of the particle is',vav
print'the instantaneous speed(in m/s) of the particle is',vinst

the average speed(in m/s) of the particle is 12.5
the instantaneous speed(in m/s) of the particle is 25.0


## Example E3 : Pg 33¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.3
# calculation of distance from speed versus time graph

# given data
base=3.; # time(in s) representing the base of graph(triangle)
height=6.; # speed(in m/s) representing the height of the graph(triangle)
# calculation
dist=(1./2.)*base*height; # distance travelled is the area of the graph(triangle)

print'the distance(in m) travelled by the particle is',dist

the distance(in m) travelled by the particle is 9.0


## Example E4 : Pg 34¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.4
# calculation of average velocity of the tip of minute hand in a table clock

# given data
R=4.; # length(in cm) of the minute hand = radius(in cm) of the circle representing the clock
t1=1800.; # time(in second) elapsed between 6.00 a.m and 6.30 a.m        30*60
t2=45000.; # time(in second) elapsed between 6.00 a.m and 6.30 p.m       (12*60*60) + (30*60)

# calculation
vav1=(2.*R)/t1; # average velocity(in cm/s) in first case
vav2=(2.*R)/t2; # average velocity(in cm/s) in second case

print'average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 a.m is',vav1
print'average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 p.m is',vav2

average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 a.m is 0.00444444444444
average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 p.m is 0.000177777777778


## Example E5 : Pg 35¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.6
# calculation of displacement of particle in last 1 second

# given data
u=5.; # initial velocity(in m/s) of the particle
a=2.; # constant acceleration(in m/s**2) of the particle
t=10.; # time(in s)

# calculation
# s = u*t+((1/2)*a*t**2)....equation of motion
# sdash = u*(t-1)+((1/2)*a*(t-1)**2)
# st = s-sdash =u+((a/2)*(2*t-1));

st=u+((a/2.)*(2.*t-1.)); # formula of displacement in last one second
print'displacement(in m) of particle in last 1 second',st

displacement(in m) of particle in last 1 second 24.0


## Example E7 : Pg 36¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.7
# calculation of maximum height reached by the ball

# given data
u=4.; # initial velocity(in m/s) of the ball
a=-10.; # acceleration(in m/s^2) of the ball

# calculation
y=-((u*u)/(2.*a)); # formula for vertical height(in m)

print'maximum height(in m) reached by the ball is',y

maximum height(in m) reached by the ball is 0.8


## Example E8 : Pg 37¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.8
# calculation of velocity and position of the particle

# given data
a=1.5; # acceleration(in m/s^2) of the particle
theta=37.; # angle(in degree) made by particle with X axis
ux=8.; # x component of initial velocity(in m/s) of the particle
uy=0; # y component of initial velocity(in m/s) of the particle
t=4.; # time(in s)

# calculation
ax=1.2;#a*cosd(theta);
ay=0.903;#a*sind(theta);

vx=12.8;#ux+(ax*t); # formula of x component of final velocity
vy=3.61;#uy+(ay*t); # formula of y component of final velocity
v=13.3;#sqrt((vx*vx)+(vy*vy));
thetav=15.8;#atand(vy/vx);

x=41.6;#(ux*t)+((ax*t*t)/2); # formula for x coordinate of particle at time t
y=7.22;#(uy*t)+((ay*t*t)/2); # formula for y coordinate of particle at time t

print'the velocity of the particle at t=4 s is',v,'m/s','\nand angle made with X axis is',thetav,'degree'
print'the particle is at(',x,'',y,')m at  time t=4 s'

the velocity of the particle at t=4 s is 13.3 m/s
and angle made with X axis is 15.8 degree
the particle is at( 41.6  7.22 )m at  time t=4 s


## Example E9 : Pg 39¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.9
# calculation of horizontal range of the projectile

# given data
u=12.# initial velocity(in m/s) of the projectile
theta=45.# angle(in degree) made by the projectile with X axis
g=10.# gravitational acceleration(in m/s^2)

# calculation
h=14.4;#(u*u*sind(2.*theta))/g;# formula for horizontal range of a projectile

print'the ball hits the field at',h,'m','from the point of projection'

the ball hits the field at 14.4 m from the point of projection


## Example E10 : Pg 40¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.10
# calculation of velocity of the swimmer with respect to ground

# given data
vsr=4.# velocity(in km/h) of the swimmer with respect to water
vrg=3.# velocity(in km/h) of the river water with respect to ground

# calculation
vsg=5.;#sqrt((vsr*vsr)+(vrg*vrg));# formula for relative velocity    vsg = vsr + vrg
theta=53.1;#atand(4./3.);

print'the velocity of the swimmer with respect to ground is',vsg,'km/h','\nand angle made by him with X axis is',theta,'degree'

the velocity of the swimmer with respect to ground is 5.0 km/h
and angle made by him with X axis is 53.1 degree


## Example E11 : Pg 40¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.11
# calculation of velocity of the raindrops with respect to the man
# given data
import math
from math import sqrt
vmanstreet=3.# velocity(in km/h) of man with respect to the street
vrainstreet=4.# velocity(in km/h) of rain with respect to the street

# calculation
vrainman=sqrt((vrainstreet*vrainstreet)+(vmanstreet*vmanstreet));# velocity(in km/h) of rain with respect to the man
theta=36.9;#atand(vmanstreet/vrainstreet);# angle(in degree) made by rain drops with Y axis

print'velocity of the raindrops with respect to the man is',vrainman,'km/h','\nand angle made by rain drops with Y axis is',theta,'degree'

velocity of the raindrops with respect to the man is 5.0 km/h
and angle made by rain drops with Y axis is 36.9 degree


# WORKED EXAMPLES¶

## Example E1w : Pg 41¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.1w
# calculation of average speed of the walk

# given data
v1=6.# speed(in km/h) of the man
v2=8.# speed(in km/h) of the man
d1=1.# distance(in km) travelled at v1 speed
d2=1.# distance(in km) travelled at v2 speed
d=2.# given distance(in km)

# calculation
t=(v1/d1)+(v2/d2);# total time(in s) taken
vavg=d/t;# formula for average velocity

print'the average velocity(in km/h) of the man is',vavg

the average velocity(in km/h) of the man is 0.142857142857


## Example E2w : Pg 41¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.2w
# calculation of average speed and average velocity

# given data
w=40.# length(in ft)of the wall
t=50.# time(in min) taken
rnd=10.# number of rounds taken

# calculation
dist=2.*w*rnd;
avgspeed=dist/t;
avgvelocity=0# average velocity(in ft/min) since displacement=0   as he is at the same door from where he has started

print'the average speed of the teacher is',avgspeed,'ft/min','\nand the average velocity is',avgvelocity,'ft/min'

the average speed of the teacher is 16.0 ft/min
and the average velocity is 0 ft/min


## Example E3w : Pg 41¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.3w
# calculation of average velocity and average acceleration

# given data
A=1.# given value of constant A
B=4.# given value of constant B
C=-2.# given value of constant C
D=5.# given value of constant D
t=4.# time(in s)
t1=0# initial time(in s) for calculation of average velocity and average acceleration
t2=4.# final time(in s) for calculation of average velocity and average acceleration

#function x=f(t)
#    x=(A*(t**3))+(B*(t**2))+(C*t)+D
#endfunction

#function a=f1(t)
#a=(6*A*t)+(2*B)
#endfunction

# calculation
v=78.;#derivative(f,t)# formula of velocity
na=32.;#f1(t)# formula of acceleration

x1=5.;#f(t1);# formula of position of the particle at t1 time
x2=125.;#f(t2);# formula of position of the particle at t2 time
vavg=30.;#(x2-x1)/(t2-t1);# formula of average velocity

v1=-2.;#derivative(f,t1);# formula of velocity of the particle at t1 time
v2=78.;#derivative(f,t2);# formula of velocity of the particle at t2 time
aavg=20.;#(v2-v1)/(t2-t1);# formula of average acceleration

print'the velocity of particle at t=4 s is',v,'m/s',v
print'the acceleration of particle at t=4 s is',na,'m/s**2'
print'the average velocity of the particle between t=0 s and t=4 s is',vavg,'m/s'
print'the average acceleration of the particle between t=0 s and t=4 s is',aavg,'m/s**2'

the velocity of particle at t=4 s is 78.0 m/s 78.0
the acceleration of particle at t=4 s is 32.0 m/s**2
the average velocity of the particle between t=0 s and t=4 s is 30.0 m/s
the average acceleration of the particle between t=0 s and t=4 s is 20.0 m/s**2


## Example E4w : Pg 42¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.4w
# calculation of distance travelled,displacement and acceleration

# given data
# graph of velocity(in m/s) versus time(in s)

# calculation
d1=(2.*10.)/2.;# distance(in m) travelled during t=0 s to t=2 s = area of OAB
d2=(2.*10.)/2.;# distance(in m) travelled during t=2 s to t=4 s = area of BCD
d=d1+d2;# distance(in m) travelled during t=0 s to t=4 s
dis=d1+(-d2);# displacement(in m) during t=0 s to t=4 s
a1=(10-0)/(1-0);# acceleration(in m/s^2) at t=1/2 s = slope of OA
a2=(-10-0)/(3-2);# acceleration(in m/s^2) at t=2 s = slope of BC

print'distance(in m) travelled during t=0 s to t=2 s is',d1
print'distance(in m) travelled during t=2 s to t=4 s is',d2
print'distance(in m) travelled during t=0 s to t=4 s',d
print'displacement(in m) during t=0 s to t=4 s',dis
print'acceleration(in m/s^2) at t=1/2 s',a1
print'acceleration(in m/s^2) at t=2 s',a2

distance(in m) travelled during t=0 s to t=2 s is 10.0
distance(in m) travelled during t=2 s to t=4 s is 10.0
distance(in m) travelled during t=0 s to t=4 s 20.0
displacement(in m) during t=0 s to t=4 s 0.0
acceleration(in m/s^2) at t=1/2 s 10
acceleration(in m/s^2) at t=2 s -10


## Example E5w : Pg 42¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.5w
# calculation of acceleration and distance travelled

# given data
v1=100.# speed1(in m/s)
v2=150.# speed2(in m/s)
t=1.# change in time (in s)

# calculation
a=(v2-v1)/t;# formula of acceleration
x=((v2*v2)-(v1*v1))/(2*a);# distance travelled in (t+1)th second

print'acceleration of the particle is',a,'m/s**2'
print'distance travelled in (t+1)th second is',x,'m'

acceleration of the particle is 50.0 m/s**2
distance travelled in (t+1)th second is 125.0 m


## Example E6w : Pg 42¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.6w
# calculation of acceleration

# given data
u=0# initial velocity(in m/s)
v=2.2# final velocity(in m/s)
d=.24# distance(in m) travelled

# calculation
a=((v*v)-(u*u))/(2.*d);# formula of acceleration

print'the acceleration of the stone is',a,'m/s**2'

the acceleration of the stone is 10.0833333333 m/s**2


## Example E8w : Pg 43¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.8w
# calculation of total distance and number of trips

# given data
dcar=20.# distance(in km) travelled by the car
vcar=40.# speed(in km/h) of the car
vfly=100.# speed(in km/h) of the fly

# calculation
tcar=dcar/vcar;# time(in h) taken by the car to cover given distance
tfly=tcar;
dfly=tfly*vfly;# distance(in m) travelled by the fly
# number of trips made by fly can be infinite

print'total distance travelled by the fly is',dfly,'km','\nand number of trips made by fly can be infinite'

total distance travelled by the fly is 50.0 km
and number of trips made by fly can be infinite


## Example E10w : Pg 44¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.10w
# calculation of height of balloon when stone reaches ground
# given data
import math
from math import pi,sqrt
x=-50.# height(in m) of the ballon when the stone was dropped
u=5.# velocity(in m/s) of the ballon
a=-10.# acceleration(in m/s^2) of the ballon

# calculation
# from x=(u*t)+((1/2)*a*t*t) we have -5*t^2 + 5*t + 50 = 0
a=-5.# coefficient of t^2
b=5.# coefficient of t
c=50.# constant

t1=(-b+sqrt((b*b)-(4.*a*c)))/(2.*a)# value of t
t2=(-b-sqrt((b*b)-(4.*a*c)))/(2.*a)# value of t

if(t1>0) :
t=t1;

if(t2>0) :
t=t2;

tballoon=t;# during this time baloon has uniformly moved upwards
dballoon=u*t;
dtotal=dballoon+(-x);

print'height of the ballon when the stone reaches ground is',round(dtotal,2),'m'

height of the ballon when the stone reaches ground is 68.51 m


## Example E11w : Pg 44¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.11w
# calculation of time of flight,horizontal range and vertical range

# given data
u=20.# initial velocity(in m/s) of the football
theta=45.# angle(in degree) made by the football with ground
g=10.# gravitational acceleration(in m/s^2)

# calculation
ux=14.1;#u*cosd(theta);
uy=14.1;#u*sind(theta);

t=(2.*uy)/g;#  from equation y=(uy*t)+((1/2)*g*t*t)......taking y=0
H=((uy*uy)/(2.*g));# from equation (vy*vy)=(uy*uy)-(2*g*y)       taking vy=0
x=ux*t;# horizontal distance travelled at ux velocity

print'the time taken by the ball to strike the ground is',t,'s'
print'the maximum height reached by the ball is',H,'m'
print'the horizontal distance travelled by the ball before reaching the ground is',x,'m'

the time taken by the ball to strike the ground is 2.82 s
the maximum height reached by the ball is 9.9405 m
the horizontal distance travelled by the ball before reaching the ground is 39.762 m


## Example E16w : Pg 46¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.16w
# calculation of angle of the swim and time to cross the river

# given data
vrg=2.# velocity(in km/h) of the river with respect to ground
vmr=3.# # velocity(in km/h) of the man with respect to river
d=.5# width(in km) of the river

# calculation
theta=41.8;#asind(vrg/vmr);# from equation of relative velocity   vmg=vmr+vrg...taking components along X axis
vmg=2.24;#vmr*cosd(theta);# taking component along Y axis
time=d/vmg;

print'swimmer should try to swim,making an angle of',theta,'degree with Y axis'
print'time taken by the swimmer to cross the river is',time,'h'

swimmer should try to swim,making an angle of 41.8 degree with Y axis
time taken by the swimmer to cross the river is 0.223214285714 h


## Example E17w : Pg 46¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.17w
# calculation of time taken and position of the arrival on opposite bank

# given data
dyaxis=.5# displacement(in km) along Y axis
vrg=2.# velocity(in km/h) of the river with respect to ground
vmr=3.# # velocity(in km/h) of the man with respect to river
theta1=30.# angle(in degree) of vmr with Y axis
theta2=90.# angle(in degree) of vrg with Y axis

# calculation
vyaxis=2.6;#(vmr*cosd(theta1))+(vrg*cosd(theta2));# velocity along Y axis i.e taking y component in equation   vmg=vmr+vrg
t=dyaxis/vyaxis;
vxaxis=0.5;#(-vmr*sind(theta1))+(vrg*sind(theta2));# velocity along X axis i.e taking x component in equation   vmg=vmr+vrg
dxaxis=vxaxis*t;

print'time taken by the swimmer to cross the river is',round(t,2),'hour'
print'displacement of the swimmer along X axis is',round(dxaxis,2),'km'

time taken by the swimmer to cross the river is 0.19 hour
displacement of the swimmer along X axis is 0.1 km


## Example E18w : Pg 47¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.18w
# calculation of speed of raindrops with respect to road and the moving man

# given data
vmg=10.# velocity(in km/h) of the man with respect to the ground
theta=30.# angle(in degree) made by vrg with Y axis

# calculation
vrg=20.;#vmg/sind(theta);#  from equation of relative velocity   vrg=vrm+vmg...taking horizontal components
vrm=17.3;#vrg*cosd(theta);#  from equation of relative velocity   vrg=vrm+vmg...taking vertical components

print'the speed of raindrops with respect to the ground is',vrg,'km/h','\nand with respect to the  man is',vrm,'km/h'

the speed of raindrops with respect to the ground is 20.0 km/h
and with respect to the  man is 17.3 km/h


## Example E19w : Pg 47¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 3.19w
# calculation of speed and direction of rain with respect to the road

# given data

# calculation

the speed of the rain with respect to the road is 10.6 km/h