In [1]:

```
# Importing module
import math
# Variable delcaration
v_r=35 # Velocity vector of the rain which falls vertically in m/s
v_w=12 # Velocity vector of the wind blowing in east to west direction in m/s
# Calculation
R=math.sqrt(v_r**2+v_w**2) # The magnitude of the resultant vector
tanθ=v_w/v_r
θ=math.degrees(math.atan(tanθ)) # The direction that R makes with the vertical
# Result
print("Therefore,the boy should hold his umbrella in the vertical plane at an angle of about",round(θ,0),"° with the vertical towards the east. ")
```

In [2]:

```
# Calculation
# Let OP and OQ represent the two vectors A and B making an angle θ.
# Then, using the parallelogram method of vector addition, OS represents the resultant vector R such that R = A + B
# SN is normal to OP and PM is normal to OS.
# From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + Bcosθ
# SN = Bsin θ and OS² = (A + Bcosθ )² + (Bsinθ )² or, R² = A² + B² + 2ABcosθ
# .i.e R = square root(A²+ B²+ 2ABcosθ)
# In ΔOSN, SN = OSsin α = Rsin α , and in ΔPSN, SN = PSsinθ = Bsinθ
# Therefore, Rsinα = Bsinθ ..............eqn 1
# Similarly, PM = Asinα = Bsinβ .........eqn 2
# From eqns 1 and 2 we get,
# sin α = (B/R) sin θ or tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)
# Result
print("The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)")
print("The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)")
```

In [3]:

```
# Importing module
import math
# Variable declaration
v_b=25 # Velocity of the motor boat racing towards north in km/h
v_c=10 # Velocity of the water current in the direction of 60° east of south in km/h
θ=60 # The angle in degree
# Calculation
# Using the parallelogram method of vector addition we can obtain the resultant vector
# We can obtain the magnitude of resultant vector using the Law of cosine
R=math.sqrt(v_b**2+v_c**2+(2*v_b*v_c*(math.cos(math.radians(2*θ)))))
# We can obtain the direction of resultant vector using the Law of sines
sinφ=(v_c*math.sin(math.radians(θ)))/R
φ=math.degrees(math.asin(sinφ))
# Result
print("The magnitude of the resultant vector =",round(R,0),"km/h")
print("The direction of the resultant vector =",round(φ,1),"°")
```

In [4]:

```
# Importing module
import math
# Variable declaration
t=1 # Time in s
# Calculation
# The position vector of the particle is r(t)=3.0ti+2.0t²j+5.0k
# On differentiating r(t) with respect to t we get the velocity vector, v(t)=3.0i+4.0tj
# On differentiating v(t) with respect to t we get the acceleration vector, a(t)=4.0j
V_x=3 # X component of v(t)
V_y=4 # Y component of v(t)
V=math.sqrt(V_x**2 + V_y**2)
tanθ=V_y/V_x
θ=math.degrees(math.atan(tanθ))
# Result
print("(a) The velocity vector, v(t) = 3.0i+4.0tj m/s")
print(" The acceleration vector, a(t) = 4.0j m/s²")
print("(b) Magnitude of v(t) at t=1 s =",V,"m/s")
print(" Direction of v(t) at t=1 s =",round(θ,0),"° with x-axis")
```

In [5]:

```
# Importing module
import sympy
# Variable declaration
x=84 # X component of the position vector in m
t= sympy.symbols('t')
# Calculation
# Velocity vector is given as, v(t)= 5.0i m/s
# Acceleration vector is given as, a(t)= 3.0i+2.0j m/s²
# By the equation r(t)=v(t)+(a(t)t²)/2 we get the positon vector of the particle as follows
# r(t) = 5.0ti + 1.5t²i + 1.0t²j
t=round((max(sympy.solve(1.5*t**2 + 5*t -x,t))),0)
y=1.0*t**2
# Now the velocity vector can be obtained by differentiating r(t) with respect to t
# Then we get, v(t) = 5.0i +3ti+2tj m/s
v_x=5.0+3*t # X component of v(t) at time = t
v_y=2*t # Y component of v(t) at time = t
V=math.sqrt(v_x**2 + v_y**2)
# Result
print("(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m =",y,"m")
print("(b) Speed of the particle at the instant its x-coordinate is 84 m =",round(V,0),"m/s")
```

In [6]:

```
# Importing module
import math
# Variable declaration
v_r=35 # Velocity vector of the rain which falls vertically in m/s
v_b=12 # Velocity vector of the bicycle riding in east to west direction in m/s
# Calculation
v_rb=v_r - v_b
tanθ=v_b/v_r
θ=math.degrees(math.atan(tanθ))
# Result
print("Therefore, the woman should hold her umbrella at an angle of about",round(θ,0),"° with the vertical towards the west.")
```

In [7]:

```
# Variable declaration
v_0=1 # For convenience, velocity at whch the projectile launched is assumed to be unity
θ_0=1 # For convenience, angle at whch the projectile launched in degree is assumed to be unity
# Result
print("For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g")
print("Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.")
print("The values of sin (90° + 2 α ) and sin (90° - 2α) are the same, equal to that of cos 2α.")
print("Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.")
```

In [8]:

```
# Importing module
import math
# Variable declaration
# We choose the origin of the x-,and y- axis at the edge of the cliff and t = 0 s at the instant the stone is thrown
# Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction
# The equations of motion are : x(t)=x_0 = v_0xt and y(t) = y_0+v_0yt+(1/2)a_y(t^2)
g=9.8 # Acceleration due to gravity
x_0=0
y_0=0
v_oy=0
a_y=g
v_ox=15
y_t=490
# Calculation
# The stone hits the ground when y(t) = 490 m ,i.e. 490 = (1/2)(9.8)t
t=math.sqrt((-y_t*-2)/a_y)
# The velocity components are v_x = v_ox and v_y = v_oy - g t
v_x=v_ox
v_y=v_oy-(g*t)
V=math.sqrt(v_x**2+v_y**2)
# Result
print("The time taken by the stone to reach the ground =",t,"s")
print("The speed with which the stone hits the ground =",round(V,0),"m/s")
```

In [9]:

```
# Importing module
import math
# Variable declaration
v_0=28 # The initial velocity of the ball in m/s
θ=30 # The angle of inclination of the ball above the horizontal in degree
g=9.8 # Acceleration due to gravity in m/s²
# Calculation
h_m=(v_0*(math.sin(math.radians(θ))))**2/(2*g)
T_f=(2*v_0*math.sin(math.radians(θ)))/g
R=((v_0**2)*(math.sin(2*math.radians(θ))))/g
# Result
print("The maximum height =",round(h_m,0),"m")
print("The time taken to return to the same level =",round(T_f,1),"s")
print("The distance from the thrower to the point where the ball returns to the same level =",round(R,0),"m")
```

In [10]:

```
# Importing module
import math
# Variable declaration
R=12 # Radius of the circular groove in cm
n=7 # Total number of revolutions
t=100 # Time taken for all 7 revolutions
# Calculation
T=t/n # Time taken for one revolution
#(a)
w=2*math.pi/T # Angular speed in rad/s
v=w*R # Linear speed in cm/s
a=pow(w,2)*R # Magnitude of acceleration in cm/s²
# Result
print("(a) Angular speed =",round(w,2),"rad/s")
print(" Linear speed =",round(v,1),"cm/s")
print("(b) Since the direction changes continuously, acceleration here is not a constant vector")
print(" Magnitude of acceleration =",round(a,1),"cm/s²")
```