Motion Summary Class 9 Science Chapter 7

Introduction

• Rest: A body is in a state of rest when its position does not change with respect to a chosen reference point.
• Motion: A body is in a state of motion when its position changes with time with respect to a chosen reference point.
• Motion can be classified by the path followed by the body:
  • Circular motion: motion along a circular path.
  • Linear motion: motion along a straight line.
  • Oscillatory (vibratory) motion: to-and-fro motion about a mean position.
• Scalar quantity: A physical quantity that has only magnitude and no direction. Example: distance, speed.
• Vector quantity: A physical quantity that has both magnitude and direction. Example: displacement, velocity, acceleration.

Distance and Displacement

• Distance: The actual length of the path travelled by an object between two positions. Distance is a scalar quantity.
• Displacement: The shortest straight-line distance from the initial position to the final position of the object, together with the direction. Displacement is a vector quantity.
• Displacement can be zero when the initial and final positions coincide (for example, in complete circular motion).

Difference between distance and displacement

• Distance is the total length of the path travelled; displacement is the straight-line change in position.
• Distance is always positive or zero; displacement can be positive, negative or zero (depending on direction and reference).
• Distance is scalar; displacement is vector.

Example 1: A body travels in a semicircular path of radius 10 m starting from point A to point B. Calculate the distance and displacement.

Distance and displacement are found as follows:

Distance travelled = length of semicircle = πR.

Substitute R = 10 m and π = 3.14.

Distance = 3.14 × 10 m = 31.4 m.

Displacement = straight-line distance between A and B = diameter = 2R = 2 × 10 m = 20 m.

Example 2: A body travels 4 km towards the North, then turns to its right and travels another 4 km before coming to rest. Calculate (i) the total distance travelled and (ii) the total displacement.

Total distance = 4 km + 4 km = 8 km.

Total displacement = straight-line distance from initial to final point. The motion forms a right-angled triangle with sides 4 km and 4 km.

Displacement magnitude = √(4² + 4²) km = √(16 + 16) km = √32 km = 4√2 km ≈ 5.657 km.

Direction: 45° east of north (i.e., northeast) from the starting point.

Uniform and Non-uniform Motion

Uniform motion

• A body has uniform motion when it covers equal distances in equal intervals of time (the speed is constant).

Non-uniform motion

• A body has non-uniform motion when it covers unequal distances in equal intervals of time (the speed is not constant).

• If the slope of the distance-time curve increases continuously, the body is accelerating (speed increasing).
• If the slope of the distance-time curve decreases continuously, the body is decelerating (speed decreasing).

Two types of non-uniform motion

• Accelerated motion: speed (or velocity) of the body increases with time.
  
• Decelerated (retarded) motion: speed (or velocity) of the body decreases with time.
  

Speed

• Speed is the measure of distance travelled per unit time.

Speed (v) = Distance travelled / Time taken = s / t

• SI unit of speed = metre per second (m s⁻¹).
• Speed is a scalar quantity; it does not include direction.
• For motion that is not uniform, the average speed over an interval is defined as:

  Average speed = Total distance travelled / Total time taken

Unit conversion: 1 km h⁻¹ = 1000 m / 3600 s = 5/18 m s⁻¹. To convert from km h⁻¹ to m s⁻¹ multiply by 5/18. To convert from m s⁻¹ to km h⁻¹ multiply by 18/5.

Example: What is the speed of a body in m s⁻¹ and km h⁻¹ if it travels 40 km in 5 hours?

Distance (s) = 40 km, Time t = 5 h.

Speed in km h⁻¹ = s / t = 40 / 5 = 8 km h⁻¹.

Convert 8 km h⁻¹ to m s⁻¹:

Speed in m s⁻¹ = 8 × (5/18) = 40/18 ≈ 2.22 m s⁻¹.

Velocity

• Velocity is speed in a given direction; it is a vector quantity.

Velocity = Displacement / Time

• SI unit of velocity = m s⁻¹.
• Average velocity over an interval = Total displacement / Total time.
• For uniformly changing velocity (constant acceleration), average velocity = (initial velocity + final velocity) / 2 = (u + v) / 2.
• Velocity can be positive, negative or zero depending on the chosen direction and motion.

Clarification (average speed vs average velocity): If equal time intervals are considered, average of speeds is (v₁ + v₂)/2. If the journey is divided into equal distances, average speed is the harmonic mean: Average speed = (2 v₁ v₂) / (v₁ + v₂) for two equal-distance parts.

Example 1: During the first half of a journey (equal distances) the body travels at 40 km h⁻¹ and in the next half at 20 km h⁻¹. Calculate the average speed of the whole journey.

For two equal distances, D/2 each, average speed = Total distance / Total time.

Time for first half = (D/2) / 40 = D / 80.

Time for second half = (D/2) / 20 = D / 40.

Total time = D/80 + D/40 = (D/80 + 2D/80) = 3D/80.

Average speed = Total distance / Total time = D / (3D/80) = 80/3 ≈ 26.67 km h⁻¹.

Example 2: A car travels 20 km in the first hour, 40 km in the second hour and 30 km in the third hour. Calculate the average speed.

Total distance = 20 + 40 + 30 = 90 km.

Total time = 1 + 1 + 1 = 3 h.

Average speed = 90 / 3 = 30 km h⁻¹.

Acceleration

• Acceleration is the rate of change of velocity with time.

Acceleration (a) = Change in velocity / Time = (v - u) / t

• SI unit of acceleration = m s⁻².
• If v > u, acceleration is positive. If v < u, acceleration is negative (deceleration or retardation).

Deceleration (retardation): When acceleration is opposite in direction to velocity, the speed decreases; acceleration value is negative.

Example 3: A car's speed increases from 40 km h⁻¹ to 60 km h⁻¹ in 5 s. Calculate its acceleration.

Convert velocities to m s⁻¹ using ×5/18.

Initial velocity u = 40 × (5/18) = 200/18 = 100/9 ≈ 11.11 m s⁻¹.

Final velocity v = 60 × (5/18) = 300/18 = 50/3 ≈ 16.67 m s⁻¹.

Time t = 5 s.

Acceleration a = (v - u) / t = (16.67 - 11.11) / 5 ≈ 5.56 / 5 ≈ 1.11 m s⁻².

Example 4: A car travelling at 20 km h⁻¹ comes to rest in 0.5 h. Calculate its retardation (in km h⁻² and m s⁻²).

Initial speed u = 20 km h⁻¹, final speed v = 0 km h⁻¹, time t = 0.5 h.

Retardation in km h⁻² = (v - u) / t = (0 - 20) / 0.5 = -40 km h⁻².

Convert to m s⁻²: 

Graphical Representation

Distance-Time (s-t) Graph

• For uniform motion the s-t graph is a straight line with constant slope. The slope gives constant speed.
  
• For non-uniform motion the s-t graph is a curve. A continuously increasing slope indicates acceleration; a continuously decreasing slope indicates deceleration.
  
  
• For a body at rest the s-t graph is a horizontal line (distance does not change with time).
  

Slope of s-t graph between two points = (s₂ - s₁) / (t₂ - t₁) = speed (or average speed in that interval).

Velocity-Time (v-t) Graph

• For uniform (constant) velocity the v-t graph is a horizontal line. Slope = 0 so acceleration = 0.
  
• For uniformly accelerated motion the v-t graph is a straight line with constant slope equal to acceleration.
  
• For non-uniformly accelerated motion the v-t graph is a curve; the slope (instantaneous rate of change of v) varies with time.
  
• For uniformly decelerated motion the v-t graph is a straight line sloping downwards at a constant rate.
  
• For non-uniform deceleration the v-t graph is a curve sloping downwards non-linearly.
  

Key facts about v-t graphs

• The slope of a v-t graph gives acceleration: a = (v₂ - v₁) / (t₂ - t₁).
• The area under a v-t curve between times t₁ and t₂ represents the displacement during that interval.

For a trapezoid or triangular region on a v-t graph:

Displacement between t₁ and t₂ = (area of triangle) + (area of rectangle) = ½ × (v₂ - v₁) × (t₂ - t₁) + v₁ × (t₂ - t₁).

Example: From the information given in an s-t graph, which body A or B is faster?

The body with the steeper slope in the s-t graph has the greater speed. Therefore v_A > v_B.

Equations of Motion (for Constant Acceleration)

When acceleration is uniform (constant), the following three equations relate displacement s, initial velocity u, final velocity v, acceleration a and time t.

First equation

v = u + at

Derivation: acceleration a = (change in velocity) / (time) = (v - u) / t. Rearranging gives v = u + at.

Second equation

s = ut + ½ a t²

Derivation from area under v-t graph: distance (displacement) in time t = area under velocity-time graph.

Area = rectangle (u × t) + triangle (½ × base t × height (v - u)).

Substitute v - u = at to obtain s = ut + ½ a t².

Third equation

v² = u² + 2as

Derivation uses elimination of t between v = u + at and s = ut + ½ a t² or geometric area methods to relate velocities and displacement. The result is v² = u² + 2as.

Example 1: A car starting from rest moves with uniform acceleration of 0.1 m s⁻² for 4 minutes. Find the final speed and distance travelled.

Given u = 0 m s⁻¹, a = 0.1 m s⁻², t = 4 min = 240 s.

Final speed v = u + at = 0 + (0.1 × 240) = 24 m s⁻¹.

Distance s = ut + ½ a t² = 0 × 240 + ½ × 0.1 × 240² = 0.05 × 57600 = 2880 m.

Example 2: Brakes produce deceleration of 6 m s⁻² opposite to motion. If a car requires 2 s to stop after application of brakes, calculate the distance travelled during this time.

Given a = -6 m s⁻², t = 2 s, final velocity v = 0 m s⁻¹.

Initial velocity u = v - at = 0 - (-6 × 2) = 12 m s⁻¹.

Distance s = ut + ½ a t² = 12 × 2 + ½ × (-6) × 2² = 24 - 12 = 12 m.

Uniform Circular Motion

• If a body moves with constant speed along a circular path, it is in uniform circular motion.
• Although the speed is constant, the direction of velocity changes continuously; therefore the motion is accelerated.
• The acceleration that keeps the body in circular motion is directed towards the centre of the circle and is called centripetal acceleration.

Centripetal acceleration magnitude (for linear speed v and radius r) is given by a_c = v² / r and is always directed towards the centre of the circular path.

Summary

• Rest and motion are defined with respect to a reference point.
• Distance and speed are scalar quantities; displacement, velocity and acceleration are vector quantities.
• For uniformly accelerated motion the three standard equations v = u + at, s = ut + ½at² and v² = u² + 2as apply.
• Graphs: slope of s-t gives speed; slope of v-t gives acceleration; area under v-t gives displacement.
• In circular motion, constant speed still implies acceleration because velocity direction changes; centripetal acceleration = v²/r.