Motion in a Plane - Lecture Notes

Motion in a Plane - Lecture 01

Topics to be Covered

1. Distance & Displacement

2. Speed & Velocity

3. Acceleration

4. Graphs

5. Rectilinear Motion

Kinematics

• Kinematics studies the motion of objects without considering the forces that cause the motion.

Distance vs. Displacement

• Distance: The actual path length traveled by a particle during its motion.

• Displacement: The change in position of a particle in a particular direction.

Example 1: Drunkard's Walk

A drunkard goes 3m to the east and 4m to the north. Find the distance and displacement.

• Distance = 3m + 4m = 7m

• Displacement:
  $AC^2 = AB^2 + BC^2$
  $AC^2 = 3^2 + 4^2 = 9 + 16 = 25$
  $AC = \sqrt{25} = 5m$

Example 2: Drunkard, Street Lamp, and Pole

A drunkard goes 3m to the east, then 6m to the north, where he finds a street lamp pole of 12m height. He climbs to the top. Find the displacement.

• Distance = 3m + 6m + 12m = 19m

• Displacement:
  First, we calculate the displacement in the 2D plane (horizontal plane):
  $AC^2 = 3^2 + 6^2 = 9 + 36 = 45$
  $AC = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \approx 6.71 m$
  Then, we consider the 3D displacement (including the vertical height):
  Let AD be the final displacement.
  $AD^2 = AC^2 + CD^2$
  $AD^2 = (3\sqrt{5})^2 + 12^2 = 45 + 144 = 189$
  $AD = \sqrt{189} = \sqrt{9 \times 21} = 3\sqrt{21} \approx 13.75 m$

Distance vs Displacement

Path length or distance: Path length is the actual distance traveled by the particle during its motion.

• Displacement: Displacement of a particle is the change in position in a particular direction.

Question:

When a car moves towards east 50m, then towards south 50m, later on towards west 50m, finally towards north 50m, the displacement of the car in magnitude is:

(A) 200m (B) 100m (C) 50m (D) zero
*Find displacement

Since the car returns to its starting point, the displacement is zero.

Question

• Find the displacement
  The height of the room is 8m. A spider starts at the bottom corner of the room and wants to catch a fly in the opposite corner at the ceiling. The room is 6m wide. If the length of the room is 10m what is the spiders displacement?

• First, consider the diagonal across the floor (2D).
  $AC^2 = 6^2 + 8^2$
  $AC^2 = 36 + 64 = 100$
  $AC = \sqrt{100} = 10$

• Now, find the 3D displacement (AD).
  $AD^2 = AC^2 + CD^2$
  $AD^2 = 10^2 + 10^2 = 100 + 100 = 200$
  $AD = \sqrt{200} = 10\sqrt{2} \approx 14.14 m$

Round Trip Examples

Consider a round trip on a circle with radius 7m.

• Full Round:

  • Distance = $2\pi R = 2 \times \frac{22}{7} \times 7 = 44 m$

  • Displacement = 0 (as it returns to the starting point)

• Half Round:

  • Distance = $\pi R = \frac{22}{7} \times 7 = 22 m$

  • Displacement = $2R = 2 \times 7 = 14 m$

• Quarter Round:

  • Distance = $\frac{1}{4} \times 2 \pi R = \frac{1}{4} \times 2 \times \frac{22}{7} \times 7 = \frac{11}{2} m$

  • Displacement = $7\sqrt{2} m$

Hypotenuse of a Right Angled Triangle

Consider a right-angled triangle with both sides equal to $x$.

• $AC^2 = x^2 + x^2 = 2x^2$

• $AC = \sqrt{2x^2} = x\sqrt{2}$

• Given values:

  • $\sqrt{2} \approx 1.414$

  • $\sqrt{3} \approx 1.73$

Speed vs. Velocity

• Average Speed = $\frac{Distance}{Time}$

• Average Velocity = $\frac{Displacement}{Time}$

Example: Sarthak's Trip to School

• Monday:

  • Sarthak goes to school (5 km away) from 7 AM to 8 AM.

  • Distance = 5 km

  • Displacement = 5 km

  • Average Speed = $\frac{5}{1} = 5 km/hr$

  • Average Velocity = $\frac{5}{1} = 5 km/hr$

• Tuesday:

  • Sarthak goes to school (5 km away) and then to Ronak's house (2 km away) from 7 AM to 9 AM.

  • Total Distance = 5 km + 2 km = 7 km

  • Displacement (from home to Ronak's house) = 5 km + 2 km = 7km

Facts

• Distance = Displacement when the body travels in the same direction.

• Average speed cannot be negative.

• Displacement/Velocity can be positive, negative, or zero.

More Example

• Sarthak goes from home to school and back home.

• Distance = 7km + 7km = 14km

• Displacement = 0

Question:

Sarthak goes 3 km away from the house and comes back 1km closer to the house.

• Displacement = -2/3 km/hr

• Thursday

  • Distance = 10m

  • Displacement = 0m

Speed and Velocity Definitions

• Average Speed: Total path length traveled during the time interval, per unit time.

• Average Velocity: Displacement of the particle during the time interval per unit time.

• Instantaneous Velocity: The limit of the average velocity as the time interval approaches zero.
  $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$

Homework

1. Find the displacement

2. Two persons complete one round in 40 seconds. Find displacement in 2 minutes 20 seconds.

3. t = 10 sec, find average speed and velocity.