GRADE 9 · KINEMATICS
Free fall, circular motion and relative velocity
Falling, centripetal acceleration and velocity relative to different observers.
Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.
Understanding step by step
A ball moves upwards while in free fall. A bicycle's speedometer reading stays constant, yet the bicycle accelerates. These statements are consistent once we keep track of the velocity vector.
Free fall as a model
Toss a ball upwards and imagine removing the air. After release, the hand no longer acts on it, but Earth still attracts it. Motion under gravity alone is called free fall. Here, fall identifies the interaction acting on the ball, not its direction of travel: the upward journey and the instant at the highest point both belong to this model. In reality, air resistance can be neglected only when that gives sufficient accuracy for the problem.
The same g
Why does a feather lag behind a stone? Imagine the experiment without air: near Earth's surface at the same location, both objects have the same downward acceleration g ≈ 9,8 м/с², regardless of mass. Air changes the result because it acts differently on the objects. During ascent, downward acceleration reduces speed; after the top, it increases the downward speed. Even when velocity is momentarily zero, acceleration remains.
Measuring acceleration
We can measure g instead of looking it up. Release an object without pushing it and time its fall through height h. The constant-acceleration equation gives h = gt²/2, so g = 2h/t². Use metres for h and seconds for t to obtain m/s². At the same g, falling from four times the height takes twice as long. Repeating the measurement shows the spread of results, but does not by itself remove air resistance.
Turning changes velocity
Now follow a cyclist around a circular track. Draw the velocity arrows at two nearby positions: each is tangent to the circle, and their directions differ. Even with the same speedometer reading, the velocity vector changes. A change in velocity over time means acceleration. If the bicycle stopped turning, it would follow its current tangent; having travelled in a circle does not make it continue around one by itself.
Why towards the centre?
In uniform circular motion, which way does the velocity change point? Place the tails of two nearby, equal-length velocity arrows together and join their tips. As the time interval shrinks, this difference points towards the centre. This gives centripetal acceleration aₙ. It is perpendicular to the current velocity: it continually changes the direction of motion without increasing speed along the path when the speed is constant.
Speed and tightness of the turn
A faster, tighter turn requires a larger change in direction each second. Circle geometry gives aₙ = v²/r, where v is speed in m/s and r is radius in metres. Doubling v makes aₙ four times as large; doubling r halves it. With a very large radius, the path is nearly straight and the turning acceleration is small. At v = 0 it is zero. The units of v²/r are m/s².
Add the carriage's motion
A passenger takes a step inside a moving carriage. Over one second, an observer on the ground sees two displacements: the carriage's displacement and the passenger's displacement inside it. The passenger's velocity relative to the ground is therefore the vector sum of their velocity relative to the carriage and the carriage's velocity relative to the ground. This is the classical rule for ordinary speeds. Along one line, choose an axis and add components; for different directions, add the vectors.
Walking against the train's motion
The train moves right at 10 m/s, and the passenger walks left through the carriage at 1 m/s. In one second the carriage carries them 10 m right, while their step subtracts 1 m: relative to the ground, 10 − 1 = 9 м/с to the right. What if they walked backwards as fast as the train moved forwards? An observer on the ground would see them at rest. Motion relative to the carriage alone does not determine motion relative to the ground.
A cyclist takes a turn
A cyclist moves at a constant speed of 6 m/s around a circle of radius 12 m. Find the acceleration.
Let's look at an example · 1
The speedometer stays at 6 m/s, but the cyclist turns. The acceleration therefore changes the direction of velocity and is centripetal.
Let's look at an example · 2
aₙ = v²/r = 6²/12 = 3 м/с². Units: (м/с)²/м = м/с².
Let's look at an example · 3
At each instant, acceleration points towards the centre, perpendicular to the tangential velocity. This does not mean the speedometer reading increases by 3 m/s every second.
Let's look at an example · 4
Halving the speed would reduce the acceleration to one-quarter, or 0,75 m/s². This checks the squared dependence.
Check the result
The acceleration is 3 m/s² towards the centre. It turns the velocity vector while keeping the speed at 6 m/s.
Key idea · 1
Free fall is motion under gravity alone; the ball may be travelling upwards at that moment.
Key idea · 2
Constant speed does not guarantee zero acceleration. On a circle, inward acceleration turns the velocity vector.
Key idea · 3
The observer's motion enters the measured velocity through vector addition.
What comes next
Next we will find out which forces cause this acceleration.
Basic Concepts
Uniform circular motion
Motion at constant speed with a continuously changing direction of the velocity vector.
Centripetal acceleration points towards the centre and changes the direction of the velocity vector.
Free fall
Motion under gravity alone, with negligible resistance from the surrounding medium.
At the same location, all freely falling objects have the same g. It can be measured from falling time and distance or from a pendulum's period, with measurement uncertainty taken into account.
Relative velocity
An object's velocity depends on the frame of reference. Its velocity in a moving frame combines with the velocity of that frame by vector addition.
The passenger's velocity relative to the ground is the vector sum of their velocity relative to the carriage and the carriage's velocity relative to the ground.
Physical quantities
Acceleration due to gravity
gAn object's acceleration when gravity is the only force acting on it.
Unit: metre per second squared · \frac{\mathrm{м}}{\mathrm{с}^2}
Radius of the circular path
rDistance from the centre of the circle to the moving object.
Unit: metre · \mathrm{м}
Centripetal acceleration
a_nAcceleration directed towards the centre of curvature of the path.
Unit: metre per second squared · \frac{\mathrm{м}}{\mathrm{с}^2}
Lesson formulas
Centripetal acceleration
a_n=\frac{v^2}{r}Centripetal acceleration equals speed squared divided by radius.
Test the relationship
Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.
Uniform circular motion
ω = v/R; x = R cos(ωt); y = R sin(ωt); a = v²/R; T = 2πR/v (v > 0)
A point particle moves counterclockwise around a circle at constant speed, starting at (R, 0) when t = 0. Arrows show directions; their lengths are schematic. This is a kinematic model: the source of force is not specified. At v = 0 the particle is at rest and the period is undefined.
At R = 2 m, speed increases from 2 to 4 m/s. How will the magnitude of acceleration change?
Initial values
Calculated result
- |v|
- 2 м/с
- a
- 2 м/с²
- T
- 6.283 с
- vₓ
- 0 м/с
- vᵧ
- 2 м/с
Explaining the result
Doubling v quadruples the acceleration magnitude, from 2 to 8 m/s². At fixed v, doubling R halves a. Velocity is tangent to the circle; acceleration points toward the centre, perpendicular to velocity. The graphs of vₓ and vᵧ change, but √(vₓ² + vᵧ²) = v stays constant. A real net force must provide this acceleration on a circular orbit.
Related formulas
The same formulas in the reference, with symbols and units.
Problems on this topic
Statements, answer options and the walkthrough.