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Physics - Speed and Velocity

Grade 7IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Speed is a scalar quantity that measures the distance covered per unit of time. It does not include direction.

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Velocity is a vector quantity that represents the rate of change of displacement. It includes both magnitude (speed) and direction (e.g., 10 m/s10 \text{ m/s} North).

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The SI unit for both speed and velocity is meters per second (m/sm/s or ms−1ms^{-1}). Other common units include km/hkm/h and mphmph.

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Average speed is the total distance traveled divided by the total time taken for a journey, accounting for variations in speed during the trip.

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Constant speed occurs when an object covers equal distances in equal intervals of time.

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On a distance-time graph, a straight diagonal line represents constant speed. The gradient (slope) of this line is equal to the speed: gradient=ΔyΔx\text{gradient} = \frac{\Delta y}{\Delta x}.

📐Formulae

v=dtv = \frac{d}{t}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

v⃗=Δs⃗Δt\vec{v} = \frac{\Delta \vec{s}}{\Delta t}

💡Examples

Problem 1:

A car travels a distance of 150 km150 \text{ km} in 2.5 hours2.5 \text{ hours}. Calculate the average speed of the car in km/hkm/h and then convert it to m/sm/s.

Solution:

Speed in km/hkm/h: v=150 km2.5 h=60 km/hv = \frac{150 \text{ km}}{2.5 \text{ h}} = 60 \text{ km/h}. To convert to m/sm/s: 60×1000 m3600 s≈16.67 m/s60 \times \frac{1000 \text{ m}}{3600 \text{ s}} \approx 16.67 \text{ m/s}.

Explanation:

We use the formula v=dtv = \frac{d}{t}. For the conversion, we multiply by 10001000 to get meters and divide by 36003600 to convert hours to seconds.

Problem 2:

A sprinter runs 100 m100 \text{ m} North in 10 seconds10 \text{ seconds}, stops, and then runs 50 m50 \text{ m} South in 5 seconds5 \text{ seconds}. Calculate the average speed and the average velocity.

Solution:

Total distance = 100+50=150 m100 + 50 = 150 \text{ m}. Total time = 10+5=15 s10 + 5 = 15 \text{ s}. Average Speed = 150 m15 s=10 m/s\frac{150 \text{ m}}{15 \text{ s}} = 10 \text{ m/s}. Total displacement = 100−50=50 m100 - 50 = 50 \text{ m} North. Average Velocity = 50 m15 s≈3.33 m/s\frac{50 \text{ m}}{15 \text{ s}} \approx 3.33 \text{ m/s} North.

Explanation:

Speed uses total distance (scalar), while velocity uses displacement (vector), which is the net change in position.

Problem 3:

Looking at a distance-time graph, a straight line starts at (0,0)(0,0) and reaches (5 s,20 m)(5 \text{ s}, 20 \text{ m}). What is the speed of the object?

Solution:

Speed=Change in distanceChange in time=20 m−0 m5 s−0 s=4 m/s\text{Speed} = \frac{\text{Change in distance}}{\text{Change in time}} = \frac{20 \text{ m} - 0 \text{ m}}{5 \text{ s} - 0 \text{ s}} = 4 \text{ m/s}.

Explanation:

The gradient of a distance-time graph represents the speed. Since the line is straight, the speed is constant.

Speed and Velocity - Revision Notes & Key Formulas | IGCSE Grade 7 Science