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Motion, Forces and Energy - Motion (Speed, velocity and acceleration)

Grade 11IGCSEPhysics

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

🔑Concepts

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Speed is a scalar quantity defined as the distance traveled per unit time, measured in m/sm/s.

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Velocity is a vector quantity defined as the displacement per unit time in a specific direction, measured in m/sm/s.

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Acceleration is the rate of change of velocity over time (m/s2m/s^2). It is a vector quantity, meaning it has both magnitude and direction.

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On a distance-time graph, the gradient (slope) represents the speed. A straight line indicates constant speed, while a horizontal line indicates the object is stationary.

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On a speed-time graph, the gradient represents the acceleration. A positive gradient indicates acceleration, a negative gradient indicates deceleration, and a horizontal line indicates constant speed.

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The area under a speed-time graph represents the total distance traveled by the object.

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Free fall: In the absence of air resistance, all objects near the Earth's surface fall with a constant acceleration called the acceleration of free fall, gg, which is approximately 9.81 m/s29.81\,m/s^2 (often rounded to 10 m/s210\,m/s^2 in IGCSE).

📐Formulae

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

a=v−uta = \frac{v - u}{t}

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

Distance (from speed-time graph)=Area under the curve\text{Distance (from speed-time graph)} = \text{Area under the curve}

💡Examples

Problem 1:

A car accelerates uniformly from a velocity of 15 m/s15\,m/s to 35 m/s35\,m/s in a time interval of 5.0 s5.0\,s. Calculate the acceleration of the car.

Solution:

a=v−ut=35 m/s−15 m/s5.0 s=20 m/s5.0 s=4.0 m/s2a = \frac{v - u}{t} = \frac{35\,m/s - 15\,m/s}{5.0\,s} = \frac{20\,m/s}{5.0\,s} = 4.0\,m/s^2

Explanation:

We use the acceleration formula where vv is the final velocity, uu is the initial velocity, and tt is the time taken.

Problem 2:

An object starts from rest and accelerates at 2 m/s22\,m/s^2 for 10 s10\,s. Determine the distance traveled by the object by sketching a speed-time graph or using the area method.

Solution:

vfinal=u+at=0+(2 m/s2×10 s)=20 m/sv_{final} = u + at = 0 + (2\,m/s^2 \times 10\,s) = 20\,m/s Distance=Area of triangle=12×base×height\text{Distance} = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} Distance=12×10 s×20 m/s=100 m\text{Distance} = \frac{1}{2} \times 10\,s \times 20\,m/s = 100\,m

Explanation:

The distance traveled is the area under the speed-time graph. Since the object starts from rest and accelerates uniformly, the graph is a triangle with base tt and height vv.

Motion (Speed, velocity and acceleration) Revision - Grade 11 Physics IGCSE