Problems on Equations of Kinematics
Q1

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 162km away. The radar antenna is pointing upward at an angle of 62.3° from the ground. Find the x and y components(in km) of the position vector of the satellite, relative to the antenna.?

Sol:

Let us assume that the radar is at the origin of the coordinate system and tracking the satellite.
Given that the satellite is found at 162km away (from the origin).
If is the position vector of the satellite, then the magnitude of the position vector is || = 162km
The angle at which the satellite is found from the ground is given, θ = 62.3°.
We know that the magnitude of the position vector in terms of its component, | | = ,
where x and y are the scalar components
⇒ = 162
⇒x2 + y2 = (162)2 ----- (A)

The angle θ that the position vector of the satellite makes with the ground, in terms of components is

θ =
⇒ 62.3° =
⇒ = 62.3°
⇒ y = tan(62.3)x
y = 1.9 x
Putting this value in equation (A), i.e.,
x2 + y2 = (162)2
x2 + (1.9) 2 x 2 = (162)2
x2[1 + 3.61] = (162) 2
Substituting this value in y, we get
y = (1.9)(75.3) = 143.2km

∴ The x and y components of the position vector of satellite are x = 75.3km, y = 143.2km

Q2

In a shopping mall, a man rides up an escalator and turns right and walks 10m to a store at the top of the escalator. The magnitude of the man's displacement from the bottom of the escalator to the store is 18m. If the escalator is inclined at an angle of 25° above the horizontal, find the vertical distance between the floors?

Sol:

Given the man's displacement towards the shop from the top of the escalator = 10m

Total displacement = 18m
Let L be the length of the escalator.
Then, from the vector addition law, we get
102 + L2 = (18)2L = 14.9m is Length of the escalator.
Now given that escalator is inclined at an angle of 25° above the horizontal

Let x be the vertical distance between the floors.

From the right angled triangle formed by the escalator, floor and the vertical distance between floor

Q3

A boy runs 85m due south for 18s. He starts from rest and stops for a negligible amount of time at the end of the run. Then he takes off again and runs 62m due east in 21s. For the entire 39s interval, Find the magnitude and direction of the boy's average velocity?

Sol:

Magnitude of north-south displacement r1 = 85m
Magnitude of east-west displacement r2 = 62m
Total time = 39s
∴ Magnitude of total displacement

∴ The magnitude of average velocity is

Now let θ be the direction of the average velocity
along the direction of displacement