102.Motion

Core + ExtendedIGCSE 0625

Motion Simulator: Distance–Time and Speed–Time Graphs

Set the initial speed and acceleration, run the car, and watch both graphs being drawn.

Each dot on the ticker tape marks where the car was at the end of every second.

Time

0.00s

Distance

0.0m

Speed

0.0m/s

Acceleration

2.0m/s²

Initial speed u0.0 m/sAcceleration a2.0 m/s²Negative values mean braking.Time t0.0 sDrag to move through the journey.

Distance–time graph

Tangent at t = 0.0 s: gradient = 0.0 m/s, the speed at that instant.

Speed–time graph

Gradient = 2.0 m/s², the acceleration. Shaded area = 0.0 m, the distance travelled.

Uniform acceleration of 2.0 m/s²: the speed rises by 2.0 m/s every second.

The speed–time graph is a straight line with gradient 2.0 m/s². The distance–time graph curves upwards, getting steeper as the car speeds up.

The dots on the ticker tape get further apart.

What is Motion?

Definition

Motion is a change in an object's position with time, measured relative to a reference point.

An object is in motion if its position changes compared with something chosen as fixed, such as the ground, a wall, or a starting line.

Exam wording

Always mention the reference point

For full marks, define motion as a change in position with time relative to a reference point or observer.

1. Distance vs Displacement

Distance is the total ground covered during motion. It ignores direction.

Displacement is the shortest straight-line distance from the starting point to the ending point.

Feature	Distance (Scalar)	Displacement (Vector)
Definition	Total path length travelled.	Shortest straight-line change in position.
Visual	Distance	Displacement
Dependencies	Depends on the path taken.	Depends ONLY on Start and End points.
Math Sign	Always Positive (+)	Positive (+) or Negative (-)
Return to Start	Distance ≠ 0 (path was covered)	Displacement = 0 (no net change)
Magnitude	Can be any value ≥ 0	Always ≤ Distance covered
Example (Track)	Running 1 lap of 400m track = 400m	Running 1 lap of 400m track = 0m
Units	metres (m), kilometres (km), etc.	metres (m) with direction (e.g., 5m North)

2. Speed vs Velocity

Speed is how fast an object is moving (magnitude only).

Velocity is speed in a given direction (magnitude + direction).

Feature	Speed (Scalar)	Velocity (Vector)
Definition	Rate of change of distance.	Rate of change of displacement.
Formula	$s = \frac{d}{t}$	$v = \frac{s}{t}$
Graph (d-t)	Distance-Time	Displacement-Time
Moving in Circle	Constant Speed (if steady pace).	Changing Velocity (direction changes constantly!).
Direction	No direction involved.	Must specify direction (e.g., 50 m/s North).
Type	Scalar quantity (magnitude only).	Vector quantity (magnitude + direction).
Reverse Motion	Always positive; doesn't show direction.	Changes sign when direction reverses.
Average Calculation	Total distance ÷ Total time	Net displacement ÷ Total time
Speedometer	Car speedometer shows speed (70 km/h).	GPS shows velocity (70 km/h Northeast).

3. Distance-Time Graphs

Stationary Object

Horizontal line = Zero gradient = Speed is 0 m/s. The object remains at a constant distance from the starting point (not moving).

Constant Speed

Straight diagonal line = Constant gradient = Constant speed. The steeper the line, the faster the speed. Distance increases uniformly over time.

Accelerating (Speeding Up)

Upward curving line = Increasing gradient = Speed is increasing. The object covers more distance in each successive time interval (acceleration).

Decelerating (Slowing Down)

Flattening curve = Decreasing gradient = Speed is decreasing. The object covers less distance in each successive time interval (deceleration).

Two Different Constant Speeds

Two straight lines with different gradients = Two different constant speeds. Steeper gradient (second part) = faster speed.

Multi-Stage Journey

Complex journey with multiple phases: (1) Accelerating (curve up), (2) Constant speed (straight), (3) Stationary (horizontal), (4) Decelerating (flattening curve).

4. Speed-Time Graphs

Constant Velocity

Horizontal line above zero axis. The speed remains the same over time. Acceleration is zero.

Constant Acceleration

Straight diagonal line sloping upwards. Speed increases by the same amount every second. The steeper the slope, the greater the acceleration.

Constant Deceleration

Straight diagonal line sloping downwards. Speed decreases by the same amount every second. Also called retardation.

Stationary (At Rest)

The line lies on the time axis (speed = 0). The object is not moving.

Increasing Acceleration

Curved line getting steeper. The rate of change of speed is increasing (e.g. car pressing throttle harder).

Decreasing Acceleration

Curved line getting flatter. The speed is still increasing, but at a slower rate (e.g. approaching max speed).

5. Acceleration and Free Fall

⚡ Definition

Acceleration is the rate at which velocity changes. It is a vector quantity.
Positive acceleration means speeding up. Negative acceleration (deceleration) means slowing down.

Mathematical Formula

a = v - ut

a = acceleration (m/s²)

v = final velocity

u = initial velocity

🌍 Real-World Free Fall

Without Air Resistance: Objects fall with constant acceleration g ≈ 9.8 m/s².

With Air Resistance: Objects reach a maximum constant speed called Terminal Velocity when air resistance balances gravity.

Visualizing Acceleration (Speed vs Time)

Adjusting Speed

Cruise Control

Overtaking

Stopping

6. Terminal Velocity

In the real world, objects falling through air experience Air Resistance (or Drag). This force opposes the motion and increases as the object moves faster.

Forces Analysis

As an object falls, the interplay between Weight (W) and Air Resistance (Drag) determines its motion. This journey happens in three key stages:

1. The Start

Velocity is zero. Only Weight acts downwards. Resultant force is maximum downwards ( $F = W$ ). Acceleration is maximum ( $g$ ).

2. Accelerating

Velocity increases. Air resistance depends on speed, so Drag increases upwards. Resultant force decreases ( $F = W - D r a g$ ). Acceleration decreases.

3. Terminal Velocity

Forces Balanced. Drag has increased until it equals Weight. Resultant force is zero ( $F = 0$ ). Acceleration is zero. Speed becomes constant.

Velocity-Time Graph

The Parachutist's Journey

Free Fall: Initially, Weight > Drag. The skydiver accelerates downwards rapidly ( $g \approx 10 m / s^{2}$ ).

Terminal Velocity 1: As speed increases, Drag increases. Eventually, Drag = Weight. Acceleration is zero and speed is constant (high).

Parachute Opens: Massive surface area increase causing Huge Drag > Weight. Rapid deceleration (slowing down).

Terminal Velocity 2: Speed drops until Drag decreases to equal Weight again. A new, lower constant speed for safe landing.

1.A sprinter accelerates from rest to 10 m/s in 2 seconds. Calculate her acceleration.

Solution

Identify knowns: u = 0 m/s, v = 10 m/s, t = 2 s.

Formula: a = (v - u) / t

Substitute: a = (10 - 0) / 2

Calculate: a = 5 m/s²

2.A car accelerates at 3 m/s² from 15 m/s to 30 m/s. How long does this take?

Solution

Identify knowns: u = 15 m/s, v = 30 m/s, a = 3 m/s².

Formula: t = (v - u) / a

Substitute: t = (30 - 15) / 3

Calculate: t = 15 / 3 = 5 s

3.A train traveling at 20 m/s accelerates at 0.5 m/s² for 30 seconds. What is its final speed?

Solution

Identify knowns: u = 20 m/s, a = 0.5 m/s², t = 30 s.

Formula: v = u + at

Substitute: v = 20 + (0.5 × 30)

Calculate: v = 20 + 15 = 35 m/s

4.A cyclist slows down from 12 m/s to 4 m/s in 4 seconds. Calculate the acceleration.

Solution

Identify knowns: u = 12 m/s, v = 4 m/s, t = 4 s.

Formula: a = (v - u) / t

Substitute: a = (4 - 12) / 4

Calculate: a = -8 / 4 = -2 m/s²

Note: Negative acceleration indicates deceleration.

5.A penny is dropped from a tall building. Calculate its speed after 3 seconds. (Take g = 9.8 m/s²)

Solution

Identify knowns: u = 0 m/s (dropped), t = 3 s, a = 9.8 m/s².

Formula: v = u + at

Substitute: v = 0 + (9.8 × 3)

Calculate: v = 29.4 m/s

6.A car accelerates at 4 m/s² for 5 seconds and reaches a speed of 30 m/s. What was its initial speed?

Solution

Identify knowns: v = 30 m/s, a = 4 m/s², t = 5 s.

Formula rearranging v = u + at gives: u = v - at

Substitute: u = 30 - (4 × 5)

Calculate: u = 30 - 20 = 10 m/s

7. Equations of Motion (SUVAT)

The SUVAT equations connect five variables of motion. They ONLY apply when acceleration is constant.

s = Displacementu = Initial Velocityv = Final Velocitya = Accelerationt = Time

📝 Derivation

Start with the definition of acceleration (gradient of v-t graph):

a =v - ut

Multiply both sides by t:

at = v - u

Rearrange to make v the subject:

v = u + at

👁️ Visual Proof

Insight: This equation comes directly from the definition of acceleration. Since acceleration is the rate of change of velocity, v increases linearly with time.

📝 Derivation

Displacement is the total Area under the v-t graph:

Area = Rectangle + Triangle

Area of the Rectangle (initial velocity component):

Area₁ = u × t = ut

Area of the Triangle (acceleration component):

Area₂ = ½ × base × height = ½ × t × (v - u)

Substitute (v - u) = at from Equation 1:

Area₂ = ½ × t × (at) = ½at²

Sum the areas to get total displacement:

s = ut + ½at²

👁️ Visual Proof

Insight: We split the area under the velocity-time graph into a rectangle (representing displacement if speed were constant) and a triangle (representing extra displacement due to acceleration).

📝 Derivation

Displacement equals Average Velocity multiplied by Time:

s = Average Velocity × t

For constant acceleration, Average Velocity is:

Avg =u + v2

Substitute Average Velocity into the displacement formula:

s =(u + v)2× t

Written in the standard form:

s = ½(u + v)t

👁️ Visual Proof

Insight: Alternatively, the area under the graph is a trapezium. The area of a trapezium is the average of the parallel sides times the width.

📝 Derivation

Rearrange Equation 1 (v = u + at) for t:

t =v - ua

Substitute this expression for t into Equation 3:

s = ½(u + v) ×(v - u)a

Simplify the numerator (Difference of Two Squares):

s =v² - u²2a

Multiply by 2a and rearrange:

v² = u² + 2as

👁️ Visual Proof

➗

Derived algebraically.

Combines Eq 1 and Eq 3 to eliminate time.

Insight: This equation is pure algebra! By combining Equation 1 and Equation 3, we eliminate 't', creating a powerful tool for solving problems where time is not known.

1.A car starts from rest and accelerates at a constant rate of 4 m/s² for 10 seconds. Calculate its final velocity.

Solution

Identify knowns: u = 0 m/s, a = 4 m/s², t = 10 s.

Identify unknown: v = ?

Choose equation: v = u + at

Substitute: v = 0 + (4 × 10)

Calculate: v = 40 m/s

2.A runner moving at 5 m/s accelerates at 2 m/s² for 4 seconds. What distance does she cover in this time?

Solution

Identify knowns: u = 5 m/s, a = 2 m/s², t = 4 s.

Identify unknown: s = ?

Choose equation: s = ut + ½at²

Substitute: s = (5 × 4) + (0.5 × 2 × 4²)

Simplify: s = 20 + (1 × 16) = 20 + 16

Calculate: s = 36 m

3.A truck traveling at 20 m/s brakes with a deceleration of 5 m/s². How far does it travel before stopping?

Solution

Identify knowns: u = 20 m/s, v = 0 m/s, a = -5 m/s².

Identify unknown: s = ?

Choose equation: v² = u² + 2as

Rearrange for s:s =v² - u²2a

Substitute:s =0² - 20²2 × (-5)

Calculate:s =-400-10= 40 m

4.A stone is dropped from a bridge. It takes 3 seconds to hit the water. Calculate the height of the bridge. (Take g = 9.8 m/s²)

Solution

Identify knowns: u = 0 m/s, t = 3 s, a = 9.8 m/s².

Identify unknown: s = ? (height)

Choose equation: s = ut + ½at²

Substitute: s = (0 × 3) + (½ × 9.8 × 3²)

Simplify: s = 0 + (4.9 × 9)

Calculate: s = 44.1 m

5.An airplane touches down at 80 m/s and decelerates at 4 m/s² until it stops. How long does this take?

Solution

Identify knowns: u = 80 m/s, v = 0 m/s, a = -4 m/s².

Identify unknown: t = ?

Choose equation: v = u + at

Rearrange for t:t =v - ua

Substitute:t =0 - 80-4

Calculate:t =-80-4= 20 s

6.A ball is thrown upwards with a velocity of 30 m/s. Calculate the maximum height it reaches. (Take g = 10 m/s²)

Solution

Identify knowns: u = 30 m/s, v = 0 m/s, a = -10 m/s².

Identify unknown: s = ?

Choose equation: v² = u² + 2as

Rearrange for s:s =v² - u²2a

Substitute:s =0² - 30²2 × (-10)

Calculate:s =-900-20= 45 m

8. Exam-Style Worked Examples

Deepen your understanding of motion with these comprehensive worked examples. Each problem breaks down the solution steps, showing you exactly how to apply formulas and interpret graphs to gain full marks in your exams.

Worked Examples

A bus travels 20 km at a speed of 40 km/h and then another 30 km at a speed of 60 km/h. Calculate the average speed for the whole journey.

Solution

Time for first part =

\frac{20}{40} = 0.5

Time for second part =

\frac{30}{60} = 0.5

Total Distance = 20 + 30 = 50 km

Total Time = 0.5 + 0.5 = 1.0 h

A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. Calculate the acceleration of the car.

Solution

Identify values:

u = 0

m/s,

v = 25

m/s,

t = 10

Use formula:

a = \frac{v - u}{t}

Substitute:

a = \frac{25 - 0}{10}

This gives

a = 2.5

m/s².

The speed-time graph for a train shows a constant speed of 20 m/s for 40 seconds, followed by a constant deceleration to rest in 10 seconds. Calculate the total distance travelled.

Solution

Distance is the area under the speed-time graph.

Area 1 (Rectangle):

20 \times 40 = 800

Area 2 (Triangle):

\frac{1}{2} \times 10 \times 20 = 100

Total Distance =

800 + 100

This gives 900 m.

A stone is dropped from a cliff. Assuming air resistance is negligible ( $g = 9.8$ m/s²), calculate how far the stone falls in 3.0 seconds.

Solution

Identify values:

u = 0

m/s,

a = 9.8

m/s²,

t = 3.0

Use formula:

s = u t + \frac{1}{2} a t^{2}

Substitute:

s = (0 \times 3) + 0.5 \times 9.8 \times (3.0)^{2}

Calculate:

s = 0 + 4.9 \times 9

This gives

s = 44.1

A distance-time graph shows a straight line starting from the origin and reaching a distance of 150 m at time 30 s. Calculate the speed of the object.

Solution

Speed is the gradient of the distance-time graph.

Gradient =

\frac{Δ y}{Δ x}

Substitute:

Speed = \frac{150}{30}

This gives 5.0 m/s.

A skydiver of mass 80 kg falls at a constant terminal velocity. Calculate the magnitude of the air resistance force acting on him. ( $g = 10$ N/kg).

Solution

At terminal velocity, acceleration is zero.

This means forces are balanced: Air Resistance = Weight.

Calculate Weight:

W = m g = 80 \times 10 = 800