This mini-lesson covers Theme A.1 — Kinematics: the language of motion (displacement, velocity, acceleration), reading motion graphs, the four equations of motion for uniform acceleration, and projectile motion.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Watch for the HL flag on higher-level extensions. Press Start when you're ready.
A.1 · quantities
Describing motion
Motion is described with vectors and scalars. Keep the pair distinct:
Distance (scalar) — total path length travelled. Displacement (vector) — straight-line change in position, with direction.
Acceleration (vector) = rate of change of velocity, a = Δv ÷ Δt, in m s⁻².
Sign convention: choose a positive direction first. A ball thrown up then falling has a constant downward acceleration of 9.81 m s⁻² the whole time — velocity changes sign, acceleration does not.
Quick check
Quick check
?A runner completes exactly one lap of a 400 m circular track, finishing where they started. What are the distance travelled and the magnitude of the displacement?
A.1 · graphs
Motion graphs
Graphs unlock a lot of marks. Two rules do most of the work:
On a displacement–time graph, the gradient is the velocity.
On a velocity–time graph, the gradient is the acceleration, and the area under the line is the displacement.
A straight, sloping line on a velocity–time graph means uniform (constant) acceleration — exactly the case the suvat equations describe.
Quick check
Quick check
?An object falling through air reaches terminal velocity. On its velocity–time graph the line becomes horizontal. What does this tell you?
A.1 · suvat
The equations of motion
For uniform acceleration in a straight line, four equations link the five quantities s, u, v, a, t. Pick the one that omits the quantity you neither know nor want.
v = u + ats = ut + ½at² · v² = u² + 2as · s = ½(u + v)t
Worked example — free fall from rest
A stone is dropped (u = 0) and falls for t = 2.5 s. Take a = g = 9.81 m s⁻².
v = u + at = 0 + 9.81 × 2.5 = 24.5 m s⁻¹
Calculate
Calculate
#A ball is dropped from rest and falls for 3.0 s. Using a = 9.81 m s⁻², find its speed just before impact (ignore air resistance).
m s⁻¹
Hint: v = u + at with u = 0, a = 9.81, t = 3.0.
Calculate
Calculate
#A car accelerates uniformly from 8.0 m s⁻¹ at 2.0 m s⁻² for 6.0 s. Find the distance travelled.
m
Hint: s = ut + ½at² = 8×6 + ½×2×6².
Calculate
Calculate
#A ball is thrown straight up at 20 m s⁻¹. Taking g = 9.81 m s⁻², find the maximum height reached (v = 0 at the top).
m
Hint: v² = u² + 2as → 0 = 20² − 2(9.81)s.
Sort it
Sort each item
Tap an item, then tap the group it belongs to.
🧭 Vector quantity
📏 Scalar quantity
📈 Read from a graph
A.1 · projectiles
Projectile motion
A projectile has independent horizontal and vertical motions, joined only by time:
Horizontal: no force (ignoring drag), so velocity is constant — use s = uₓt.
Vertical: constant acceleration g downward — use the suvat equations.
Worked example — horizontal launch
A ball rolls off a bench and lands after falling h = 1.25 m. Time to fall:
t = √(2h ÷ g) = √(2 × 1.25 ÷ 9.81) = 0.505 s
Calculate
Calculate
#A stone is thrown horizontally from a 45 m high cliff. Using g = 9.81 m s⁻², how long does it take to hit the ground below?
s
Hint: vertical only — h = ½gt², so t = √(2h ÷ g) = √(90 ÷ 9.81).
Match it
Match the situation to the right equation
Tap a statement on the left, then its match on the right.
Statement
Answer
A.1 · HL depth
Launch at an angle
When a projectile is launched at angle θ to the horizontal at speed u, resolve first:
Horizontal component uₓ = u cos θ (constant).
Vertical component u_y = u sin θ (decreases at g).
Time of flight over level ground: T = 2u_y ÷ g. Range: R = uₓ × T. Maximum height uses v² = u² + 2as with the vertical component.
The path is a parabola. The maximum range over level ground occurs at θ = 45°, where the horizontal and vertical components are balanced.
Calculate
Calculate
#HL: a ball is launched at 25 m s⁻¹ at 30° above the horizontal over level ground (g = 9.81 m s⁻²). Find the horizontal range.
m
Hint: u_y = 25 sin30 = 12.5; T = 2u_y ÷ g; range = 25 cos30 × T.
Recap
The big ideas to know
Vectors vs scalars: displacement/velocity/acceleration carry direction; distance/speed/time do not
Graphs: v–t gradient = acceleration; area under v–t = displacement
Equations of motion: v = u+at · s = ut+½at² · v² = u²+2as · s = ½(u+v)t
Projectiles: independent horizontal (constant v) and vertical (accelerate at g) motions
HL: launch at angle: resolve into components, T = 2u sinθ ÷ g
That completes Kinematics for IB Diploma Physics HL. Press Finish to see your score.
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