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📚 Scalar and Vector Quantities
A scalar quantity has magnitude only — just a size or amount. A vector quantity has both magnitude and direction.
Vectors are represented by arrows. The length of the arrow shows the magnitude; the direction of the arrow shows the direction of the quantity.
Key Examples
ForceA vector — always needs a direction. "50 N upwards" is correct. "50 N" alone is incomplete.
MassA scalar — just a quantity of matter in kg. It has no direction. Never say "75 kg downwards".
AccelerationA vector — it is rate of change of velocity, which is itself a vector. Direction is inherited.
Energy & PowerScalars — they are amounts/rates with no directional meaning.
Q1Define a vector quantity.
1 mark
Model Answer
A vector quantity has both magnitude (size) AND direction.
1 mark
Q2Define a scalar quantity.
1 mark
Model Answer
A scalar quantity has magnitude (size) only — it has no direction.
1 mark
Q3Give three examples of vector quantities.
1 mark
Model Answer
Any three from: displacement, velocity, acceleration, force, weight, momentum
All are quantities that have both size and direction
1 mark
Q4Give three examples of scalar quantities.
1 mark
Model Answer
Any three from: distance, speed, mass, energy, time, temperature, power
1 mark
Q5James says "the force of air resistance on the parachute is 50N". What else should James say to be completely correct? Why?
2 marks
Model Answer
He should state the direction — e.g. "50N upwards" (1)
Force is a vector quantity, so a complete description must include both magnitude and direction (1)
2 marks
Q6Molly says "the mass of the parachutist is 75kg towards the centre of the Earth". What is wrong with this statement? Why?
2 marks
Model Answer
Mass is a scalar quantity — it has no direction (1)
Saying "towards the centre of the Earth" implies a direction, which is meaningless for mass. She may be confusing mass with weight, which is a vector force (1)
2 marks
Q7Look at the skier diagram. Name the vector quantities shown.
1 mark
Model Answer
Normal force (perpendicular to slope), air resistance (opposing motion), weight (downward), friction (opposing motion along slope)
All four are forces — forces are always vector quantities
1 mark
Q8"Energy" is a scalar quantity. Explain why.
1 mark
Model Answer
Energy has no associated direction — you cannot say energy acts "northward" or "upward". It is simply an amount stored or transferred, making it scalar.
1 mark
Q9"Acceleration" is defined as "rate of change of velocity". Is acceleration a scalar or vector quantity? Explain why.
2 marks
Model Answer
Acceleration is a vector quantity (1)
Because velocity is a vector (has direction), any change in velocity also has a direction. Acceleration inherits this directional property from velocity (1)
2 marks
Q10"Power" is defined as "rate of transfer of energy". Is power a scalar or vector quantity? Explain why.
2 marks
Model Answer
Power is a scalar quantity (1)
Because energy is scalar (has no direction), the rate of transfer of energy also has no direction. Power therefore has magnitude only, making it scalar (1)
2 marks
📚 Distance and Displacement
DistanceA scalar — the total length of path travelled regardless of direction. Always positive.
DisplacementA vector — the straight-line distance from start to finish, including direction.
Displacement can be equal to distance (if you travel in a straight line without changing direction) but it can never be greater than the distance travelled. If you return towards your starting point, displacement decreases while distance keeps increasing.
Number Line Example
A (0 km) → B (5 km) → C (9 km) → back to B (5 km): Distance = 5 + 4 + 4 = 13 km. Displacement = 5 km right (ends at B).
Q1Describe the difference between distance and displacement.
2 marks
Model Answer
Distance is the total length of path travelled by an object — it is a scalar quantity with magnitude only (1)
Displacement is the straight-line distance from start to finish, including direction — it is a vector quantity (1)
2 marks
Q2Draw a diagram to show the difference between distance and displacement.
2 marks
Model Answer
Diagram should show: a curved/zigzag path (total distance) between two points, and a straight arrow from start to finish labelled displacement (1)
Displacement arrow is shorter than or equal to the total distance path (1)
The displacement is always the straight-line "as the crow flies" distance from start to finish
2 marks
Q3Which quantity is a vector quantity: distance or displacement?
1 mark
Model Answer
Displacement — it has both magnitude and direction (e.g. 6.5m to the right)
1 mark
Q4True or false: For a single object in motion, displacement can be equal to distance but it can never be bigger.
1 mark
Model Answer
True — displacement equals distance only when the object travels in a perfectly straight line in one direction. For any curved or return path, displacement will be less than distance. Displacement can never exceed distance.
1 mark
Q5A car starts at A (0km) and travels to B (5km), then to C (9km), and then back to B (5km). What is: a) the distance travelled and b) the displacement of the car?
2 marks
Model Answer
a) Distance = AB + BC + CB = 5 + 4 + 4 = 13 km (1)
b) Displacement = final position − start = 5 − 0 = 5 km to the right (ends at B) (1)
2 marks
Q6a) What is the displacement from B to A? b) What is the distance from B to A?
2 marks
Model Answer
a) Displacement = 5 km to the left (same magnitude as A→B but in the opposite direction) (1)
b) Distance = 5 km (the straight-line path length from B back to A) (1)
Displacement has direction — returning reverses the sign/direction
2 marks
ExtExtension: A walker travels A→B→C→D→E→F on the grid (scale 0.5km). What is the distance? What is the displacement?
2 marks
Model Answer
Count each segment on the grid using 0.5km scale, sum for total distance (1)
Displacement = straight-line distance from A to F with direction (significantly less than total distance) (1)
Use Pythagoras on the overall horizontal/vertical components to find straight-line displacement from A to F
2 marks
📚 Speed and Velocity
Speed = how much distance is covered per unit time — a scalar.
Velocity = speed in a given direction — a vector. It is the rate of change of displacement. Always include direction in your answer (e.g. "2.5 m/s to the right").
v = d / t | d = v × t | t = d / v
VESSU Method
V — ValuesWrite out all known values: d = ?, v = ?, t = ?
E — EquationWrite the equation: v = d / t
S — SubstituteReplace letters with numbers
S — SolveCalculate the answer
U — UnitsWrite units: m/s (and direction for velocity)
Finding Velocity — v = d / t
Q1What is the difference between speed and velocity?
2 marks
Model Answer
Speed is a scalar quantity — it only has magnitude (e.g. 14 m/s) (1)
Velocity is a vector quantity — it has both magnitude and direction (e.g. 14 m/s north) (1)
2 marks
Velocity calculations — v = d / t
Q2Find the velocity of a car which travels 25m to the right over 10 seconds.
2 marks
Model Answer
v = d / t = 25 / 10 = 2.5 m/s to the right
Always state direction for velocity
2 marks
Q3What is the velocity of a hot air balloon which travels 125m up in 50 seconds?
2 marks
Model Answer
v = d / t = 125 / 50 = 2.5 m/s upwards
2 marks
Q4A diver travels 640m down during a 16-second dive. Find his velocity.
2 marks
Model Answer
v = 640 / 16 = 40 m/s downwards
2 marks
Q11Find the velocity of a car which travels 35m to the right over 40 seconds.
2 marks
Model Answer
v = 35 / 40 = 0.875 m/s to the right
2 marks
Q12What is the velocity of a hot air balloon which travels 325m up in 59 seconds?
2 marks
Model Answer
v = 325 / 59 = 5.51 m/s upwards
2 marks
Q13A diver travels 645m down during a 12-second dive. Find his velocity.
2 marks
Model Answer
v = 645 / 12 = 53.75 m/s downwards
2 marks
Displacement calculations — d = v × t
Q5What is the displacement of a rocket travelling at 150m/s up for 30 seconds?
2 marks
Model Answer
d = v × t = 150 × 30 = 4,500 m upwards
2 marks
Q6Find the displacement of a beetle crawling for 180 seconds at 0.25m/s to the left.
2 marks
Model Answer
d = 0.25 × 180 = 45 m to the left
2 marks
Q7A car has a velocity of 24m/s to the right and travels for 58 seconds. Find its displacement.
2 marks
Model Answer
d = 24 × 58 = 1,392 m to the right
2 marks
Q14What is the displacement of a rocket travelling at 350m/s up for 20 seconds?
2 marks
Model Answer
d = 350 × 20 = 7,000 m upwards
2 marks
Q15Find the displacement of a beetle crawling for 380 seconds at 0.75m/s to the left.
2 marks
Model Answer
d = 0.75 × 380 = 285 m to the left
2 marks
Q16A car has a velocity of 34m/s to the right and travels for 48 seconds. Find its displacement.
2 marks
Model Answer
d = 34 × 48 = 1,632 m to the right
2 marks
Time calculations — t = d / v
Q8A car travelling at 2.75m/s travels 185m. What time was taken?
2 marks
Model Answer
t = d / v = 185 / 2.75 = 67.3 s
2 marks
Q9Find the time taken by a rocket travelling at 240m/s to reach a displacement of 15,000m.
2 marks
Model Answer
t = 15,000 / 240 = 62.5 s
2 marks
Q10A snail travels at 0.08m/s and reaches a displacement of 0.2m. Find the time taken.
2 marks
Model Answer
t = 0.2 / 0.08 = 2.5 s
2 marks
Q17A car travelling at 3.75m/s travels 196m. What time was taken?
2 marks
Model Answer
t = 196 / 3.75 = 52.3 s
2 marks
Q18Find the time taken by a rocket travelling at 170m/s to reach a displacement of 46,000m.
2 marks
Model Answer
t = 46,000 / 170 = 270.6 s
2 marks
Q19A snail travels at 0.02m/s and reaches a displacement of 0.4m. Find the time taken.
2 marks
Model Answer
t = 0.4 / 0.02 = 20 s
2 marks
📚 Distance-Time Graphs — Key Rules
Reading the Gradient
The gradient (slope) of a distance-time graph equals the speed. To find speed: choose two points on the line, then:
Speed = Δdistance / Δtime = rise / run
Horizontal lineSpeed = 0 — stationary.
Straight lineConstant speed — steeper = faster.
Curved lineChanging speed — accelerating or decelerating.
Drawing Lines
To draw 60m in 5s: plot (0, 0) and (5, 60), draw straight line. Gradient = 60/5 = 12 m/s. Higher starting point on y-axis just means the object started further away — gradient unchanged.
IntroWhat does the gradient of a distance-time graph represent?
1 mark
Model Answer
The gradient of a distance-time graph equals the speed of the object.
A steeper gradient = higher speed. A horizontal line = stationary. A curved line = changing speed (acceleration).
1 mark
Drawing lines on distance-time graphs — pp.9–10
Q1p9Draw lines to show: a) 60m in 5s, b) 100m in 5s, c) 20m in 5s. Then find the speed for each.
3 marks
Model Answer
a) Line from (0,0) to (5,60). Speed = 60/5 = 12 m/s b) Line from (0,0) to (5,100). Speed = 100/5 = 20 m/s c) Line from (0,0) to (5,20). Speed = 20/5 = 4 m/s
Gradient = rise/run = distance/time = speed
3 marks
Q2p10Draw lines for: a) 10m in 10s, b) 100m in 10s, c) 40m in 10s. Find the speeds.
3 marks
Model Answer
a) Line (0,0)→(10,10). Speed = 10/10 = 1 m/s b) Line (0,0)→(10,100). Speed = 100/10 = 10 m/s c) Line (0,0)→(10,40). Speed = 40/10 = 4 m/s
3 marks
Reading speed from distance-time graphs — p.11
Q1p11Find the speed for each graph (a–f): a) 3m in 3s, b) 3m in 4s, c) 4m in 2s, d) 8m in 2s, e) 8m in 8s, f) 12m in 8s.
6 marks
Model Answer
a) 3/3 = 1 m/s b) 3/4 = 0.75 m/s c) 4/2 = 2 m/s d) 8/2 = 4 m/s e) 8/8 = 1 m/s f) 12/8 = 1.5 m/s
Speed = Δdistance / Δtime = gradient of the line
6 marks
Q2p11Find the speed for each graph (a–c): a) 20m in 8s, b) 6m in 4s, c) 240m in 12s.
3 marks
Model Answer
a) 20/8 = 2.5 m/s b) 6/4 = 1.5 m/s c) 240/12 = 20 m/s
DT_curvesWhat does a curved line on a distance-time graph mean?
2 marks
Model Answer
A curved line means the gradient is changing — i.e. the speed is changing — so the object is accelerating or decelerating (1)
A curve getting steeper = increasing speed (positive acceleration). A curve getting flatter = decreasing speed (negative acceleration/deceleration) (1)
2 marks
Q1p14Graph 1 (curve starts steep then flattens, d increasing): Positive or negative acceleration? Moving forwards or backwards?
2 marks
Model Answer
Negative acceleration (deceleration) — the gradient is decreasing (flattening) (1)
Moving forwards — distance is increasing (1)
2 marks
Q2p14Graph 2 (curve starts steep going upward, then flattens at high d): Positive or negative acceleration? Forwards or backwards?
2 marks
Model Answer
Negative acceleration (deceleration) (1)
Moving backwards — the distance from the origin is decreasing as the object slows (1)
2 marks
Q3p14Graph 3 (d starts high and decreases steeply, curve getting steeper going down): Positive or negative? Forwards or backwards?
2 marks
Model Answer
Positive acceleration (1)
Moving backwards — distance is decreasing and the magnitude of the gradient is increasing (speeding up in the backward direction) (1)
2 marks
Q4p14Graph 4 (starts at 0, curves upward getting steeper): Positive or negative acceleration? Forwards or backwards?
2 marks
Model Answer
Positive acceleration (1)
Moving forwards — distance is increasing and the gradient is getting steeper (speeding up) (1)
2 marks
Q5p14How can we find the momentary speed of an accelerating object from a distance-time graph?
1 mark
Model Answer
Draw a tangent to the curve at the required time, then calculate the gradient of that tangent line. Gradient = speed at that instant.
1 mark
Q6p14Using the tangent method on the given graph: find the speed at a) 7s, b) 11s, c) 13s. Is the object moving forwards or backwards? Accelerating or decelerating?
5 marks
Model Answer
Draw tangent at each time point and calculate gradient = Δd/Δt a) At 7s: tangent gradient ≈ read from your graph (example: ~4 m/s) b) At 11s: tangent gradient ≈ ~8 m/s (steeper) c) At 13s: tangent gradient ≈ ~14 m/s (steeper still) d) Forwards — distance is increasing throughout e) Accelerating — gradient is getting steeper over time
Exact values depend on reading your own tangent lines carefully from the graph
5 marks
📚 Curved Distance-Time Graphs and the Tangent Method
Curve going upwardMoving forward (distance from start increasing).
Curve going downwardMoving backward (distance from start decreasing).
Finding Speed at a Point — Draw a Tangent
A tangent is a straight line that just touches the curve at one point. Draw it as accurately as possible, then calculate its gradient:
Example from booklet: at t = 10s, tangent gives Δd = 40m over Δt = 7.5s → speed = 40/7.5 = 5.33 m/s.
Q7p14Graph 7: Curved line that rises then levels off. Find speed at a) 7s, b) 10s, c) 12s, d) 14s. Forwards/backwards? Accelerating/decelerating?
6 marks
Model Answer
Draw tangents at each point and read gradient a) At 7s: tangent ≈ steep positive (e.g. ~8 m/s) b) At 10s: less steep (e.g. ~4 m/s) c) At 12s: very shallow (e.g. ~2 m/s) d) At 14s: nearly flat (e.g. ~0.5 m/s) e) Forwards — distance still increasing throughout f) Decelerating — gradient is decreasing (getting flatter)
The gradient decreases over time = negative acceleration
6 marks
📚 What is Acceleration?
Acceleration is the rate of change of velocity. Since velocity is a vector (has direction), any change in velocity counts as acceleration — including:
Changing directionAcceleration even at constant speed — direction of velocity changes, e.g. turning a corner, orbiting.
BouncingAt the moment of bounce, velocity changes from downward to upward — large acceleration.
An acceleration always requires a resultant (net) force to cause it (Newton's Second Law: F = ma).
Q1Define "acceleration".
1 mark
Model Answer
Acceleration is the rate of change of velocity.
1 mark
Q2What is the unit for acceleration?
1 mark
Model Answer
Metres per second squared (m/s²)
1 mark
Q3Explain why applying the brakes in a moving car causes it to accelerate.
2 marks
Model Answer
Acceleration is any change in velocity (1)
Applying the brakes causes the velocity to decrease — this change in velocity (from higher to lower, i.e. a negative change) is a form of acceleration called deceleration (1)
2 marks
Q4Explain why a pebble landing on the ground undergoes acceleration.
2 marks
Model Answer
Acceleration is any change in velocity (1)
The pebble's velocity changes suddenly from downward to zero when it hits the ground — this rapid change in velocity means it undergoes (very large) acceleration (1)
2 marks
Q5Explain why a bouncing ball is accelerating.
2 marks
Model Answer
Acceleration is any change in velocity (1)
At the moment of the bounce, the velocity changes direction from downward to upward — this is a change in velocity, therefore an acceleration occurs (1)
2 marks
Q6Explain why the Moon orbiting the Earth is accelerating.
2 marks
Model Answer
Velocity is speed with direction (1)
Although the Moon moves at roughly constant speed, its direction continuously changes as it orbits. Since direction is part of velocity, the velocity is constantly changing — this constitutes acceleration (centripetal acceleration) (1)
2 marks
Q7What is the relationship between force and acceleration?
2 marks
Model Answer
They are proportional — the greater the resultant force, the greater the acceleration (F = ma) (1)
They act in the same direction — the acceleration is in the direction of the resultant force (1)
2 marks
Acceleration formula: a = Δv / t — pp.18–19
Q3An object accelerates from 5m/s to 10m/s over 20s. Find the acceleration.
2 marks
Model Answer
Δv = 10 − 5 = 5 m/s a = Δv / t = 5 / 20 = 0.25 m/s²
2 marks
Q4Find the acceleration of an object starting at 0.2m/s increasing to 12.5m/s over 5s.
2 marks
Model Answer
Δv = 12.5 − 0.2 = 12.3 m/s a = 12.3 / 5 = 2.46 m/s²
2 marks
Q5What acceleration takes an object from 20m/s to 8m/s over 3s?
2 marks
Model Answer
Δv = 8 − 20 = −12 m/s a = −12 / 3 = −4 m/s² (deceleration = 4 m/s²)
2 marks
Q6An object decelerates from 100m/s to 75m/s over 5s. Calculate the deceleration.
2 marks
Model Answer
Δv = 75 − 100 = −25 m/s a = −25 / 5 = −5 m/s²
2 marks
Q7Object decelerates from 45m/s to 13m/s over 12s.
2 marks
Model Answer
Δv = 13 − 45 = −32 m/s a = −32 / 12 = −2.67 m/s²
2 marks
Q8Object decelerates from 48m/s to 3m/s over 30s.
2 marks
Model Answer
Δv = 3 − 48 = −45 m/s a = −45 / 30 = −1.5 m/s²
2 marks
Q9Object starts at rest and accelerates to 14m/s over 3.5s.
2 marks
Model Answer
Δv = 14 − 0 = 14 m/s a = 14 / 3.5 = 4 m/s²
2 marks
Q10Object accelerates from rest to 105m/s in 55s.
2 marks
Model Answer
a = 105 / 55 = 1.91 m/s²
2 marks
Q11What acceleration takes an object from rest to 64m/s in 4s?
2 marks
Model Answer
a = 64 / 4 = 16 m/s²
2 marks
Q12Object at 35m/s slows to stop over 70s.
2 marks
Model Answer
Δv = 0 − 35 = −35 a = −35 / 70 = −0.5 m/s²
2 marks
Q13Object decelerates from 12m/s to rest over 2.5s.
2 marks
Model Answer
a = −12 / 2.5 = −4.8 m/s²
2 marks
Q14Object at 58m/s stops over 4s.
2 marks
Model Answer
a = −58 / 4 = −14.5 m/s²
2 marks
Finding time — t = Δv / a
Q15How long would it take an object with deceleration 2m/s² to go from 10m/s to 4m/s?
2 marks
Model Answer
Δv = 4 − 10 = −6 m/s t = Δv / a = −6 / −2 = 3 s
2 marks
Q16Time for an object with acceleration 4.3m/s² to go from 9m/s to 85m/s?
2 marks
Model Answer
Δv = 85 − 9 = 76 m/s t = 76 / 4.3 = 17.67 s
2 marks
Q17Object with acceleration 0.45m/s², time from rest to 3m/s.
2 marks
Model Answer
t = 3 / 0.45 = 6.67 s
2 marks
Finding change in velocity — Δv = a × t
Q18Acceleration 6m/s², what is Δv over 15s?
2 marks
Model Answer
Δv = 6 × 15 = +90 m/s
2 marks
Q19Acceleration −18m/s², what is Δv over 76s?
2 marks
Model Answer
Δv = −18 × 76 = −1,368 m/s
2 marks
Q20Acceleration 0.9m/s², Δv over 200s?
2 marks
Model Answer
Δv = 0.9 × 200 = +180 m/s
2 marks
📚 Calculating Acceleration — a = Δv / t
a = (v − u) / t where v = final velocity, u = initial velocity
Positive resultObject is speeding up.
Negative resultObject is decelerating (slowing down).
u = 0Object starts from rest — initial velocity is zero.
v = 0Object comes to a stop — final velocity is zero.
Rearrangements
v = u + at | t = Δv / a | Δv = a × t
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📚 Velocity-Time Graphs — Reading and Interpreting
Do not confuse velocity-time graphs with distance-time graphs! On a v-t graph:
y-axisVelocity (m/s) — not distance.
Horizontal lineConstant velocity (could be moving at steady speed, or stationary if v=0).
Slope upwardAcceleration — velocity is increasing.
Slope downwardDeceleration — velocity is decreasing.
Area under graphDistance (displacement) travelled.
IntroWhat does the gradient of a velocity-time graph represent?
1 mark
Model Answer
The gradient of a velocity-time graph equals the acceleration of the object.
A positive gradient = positive acceleration. A negative gradient = deceleration. Zero gradient (horizontal) = constant velocity.
1 mark
Identifying graph types — p.21
VT1State what each graph shows — Graph 1: horizontal line at v=4.
1 mark
Model Answer
Constant velocity (steady speed, zero acceleration)
1 mark
VT2Graph 2: line sloping upward from 0 to 8 m/s over 10s.
Stationary (object not moving — velocity = 0 throughout)
1 mark
Finding acceleration from v-t graph gradient — pp.22–25
VT1Find the acceleration from V-T graph 1: 0→30 in 8s.
2 marks
Model Answer
Δv = 30 − 0 = 30 m/s Δt = 8 − 0 = 8 s a = Δv / Δt = 30/8 = 3.75 m/s²
2 marks
VT2Find the acceleration from V-T graph 2: 10→20 in 8s.
2 marks
Model Answer
Δv = 20 − 10 = 10 m/s Δt = 8 − 0 = 8 s a = Δv / Δt = (20−10)/8 = 1.25 m/s²
2 marks
VT3Find the acceleration from V-T graph 3: 30→15 in 8s.
2 marks
Model Answer
Δv = 15 − 30 = -15 m/s Δt = 8 − 0 = 8 s a = Δv / Δt = (15−30)/8 = −1.875 m/s²
2 marks
VT4Find the acceleration from V-T graph 4: 20→0 in 4s.
2 marks
Model Answer
Δv = 0 − 20 = -20 m/s Δt = 4 − 0 = 4 s a = Δv / Δt = (0−20)/4 = −5 m/s²
2 marks
VT5Find the acceleration from V-T graph 5: 10→20 in 8s.
2 marks
Model Answer
Δv = 20 − 10 = 10 m/s Δt = 8 − 0 = 8 s a = Δv / Δt = 10/8 = 1.25 m/s²
2 marks
VT6Find the acceleration from V-T graph 6: 30→0 in 6s.
2 marks
Model Answer
Δv = 0 − 30 = -30 m/s Δt = 6 − 0 = 6 s a = Δv / Δt = (0−30)/6 = −5 m/s²
2 marks
VT7Find the acceleration from V-T graph 7: 10→30 in 4s.
2 marks
Model Answer
Δv = 30 − 10 = 20 m/s Δt = 4 − 0 = 4 s a = Δv / Δt = 20/4 = 5 m/s²
2 marks
VT8Find the acceleration from V-T graph 8: 30→10 in 8s.
2 marks
Model Answer
Δv = 10 − 30 = -20 m/s Δt = 8 − 0 = 8 s a = Δv / Δt = (10−30)/8 = −2.5 m/s²
2 marks
📚 Finding Acceleration from a Velocity-Time Graph
The gradient of a velocity-time graph equals the acceleration:
a = Δv / Δt = (v₂ − v₁) / (t₂ − t₁)
Important: use the change in velocity, not the final velocity. If the line starts at v = 2 m/s and ends at v = 8 m/s over 10 s: Δv = 8 − 2 = 6 m/s, a = 6/10 = 0.6 m/s²
If the gradient is negative (line slopes downward), the acceleration is negative:
E.g. v goes 6→0 in 10s: Δv = 0−6 = −6, a = −6/10 = −0.6 m/s²
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📚 v² = u² + 2as — When and How to Use It
This equation links velocity, acceleration and displacement — use it when time is not given and not needed.
Step-by-Step Method (VESSU)
1. ValuesList v, u, a, s — write "?" for the unknown.
2. EquationWrite v² = u² + 2as
3. SubstituteReplace letters with numbers.
4. SolveRearrange for the unknown (see formula card above).
5. Unitsm/s for velocity, m/s² for acceleration, m for displacement.
Common Situations
u = 0Object starts from rest — u² = 0, so v² = 2as
v = 0Object stops — 0 = u² + 2as, so u² = −2as
DecelerationAcceleration a is negative — substitute as a negative number.
IntroState the equation v² = u² + 2as and define each variable.
1 mark
Model Answer
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
s = displacement (m)
Use this formula when time is not known.
1 mark
Section A & B — Finding values, substituting — pp.26–27
A1Find the final velocity: u=9m/s, a=7.8m/s², s=80m.
560² = u² + 2×0.8×2,500 313,600 = u² + 4,000 u² = 309,600 u = √309,600 = 556.4 m/s
3 marks
B4v=10, u=3, a=2.5 — find s.
3 marks
Model Answer
10² = 3² + 2×2.5×s 100 = 9 + 5s 91 = 5s s = 18.2 m
3 marks
B5u=4, a=3.8, s=31 — find v.
3 marks
Model Answer
v² = 16 + 2×3.8×31 = 16 + 235.6 = 251.6 v = √251.6 = 15.86 m/s
3 marks
B6v=23, a=4.1, s=100 — find u.
3 marks
Model Answer
23² = u² + 2×4.1×100 529 = u² + 820 u² = 529 − 820 = −291 → This gives a negative u² which is mathematically impossible. Check values: the object cannot reach v=23m/s from this starting point with a=4.1 over 100m — likely a booklet error.
3 marks
Section C — Finding v — pp.27–28
C1u=2, a=30, s=45 — find v.
3 marks
Model Answer
v² = 4 + 2×30×45 = 4 + 2,700 = 2,704 v = √2,704 = 52 m/s
3 marks
C2u=40, a=6.8, s=230 — find v.
3 marks
Model Answer
v² = 1,600 + 2×6.8×230 = 1,600 + 3,128 = 4,728 v = √4,728 = 68.8 m/s
3 marks
C3u=650, a=1,300, s=490 — find v.
3 marks
Model Answer
v² = 422,500 + 2×1,300×490 = 422,500 + 1,274,000 = 1,696,500 v = 1,302.5 m/s
3 marks
C4u=3, a=4.8, s=24 — find v.
3 marks
Model Answer
v² = 9 + 230.4 = 239.4 v = 15.47 m/s
3 marks
C5u=25, a=0.5, s=200 — find v.
3 marks
Model Answer
v² = 625 + 200 = 825 v = 28.72 m/s
3 marks
C6u=0.2, a=1.9, s=140 — find v.
3 marks
Model Answer
v² = 0.04 + 532 = 532.04 v = 23.07 m/s
3 marks
Section D — Finding u — pp.29
D7v=135.7, a=22.1, s=100 — find u.
3 marks
Model Answer
135.7² = u² + 2×22.1×100 18,414.5 = u² + 4,420 u² = 13,994.5 u = 118.3 m/s
3 marks
D8v=655, a=25, s=239 — find u.
3 marks
Model Answer
655² = u² + 2×25×239 429,025 = u² + 11,950 u² = 417,075 u = 645.8 m/s
3 marks
D9v=0.64, a=0.21, s=0.31 — find u.
3 marks
Model Answer
0.64² = u² + 2×0.21×0.31 0.4096 = u² + 0.1302 u² = 0.2794 u = 0.529 m/s
3 marks
D10v=207, a=3.1, s=700 — find u.
3 marks
Model Answer
207² = u² + 2×3.1×700 42,849 = u² + 4,340 u² = 38,509 u = 196.2 m/s
3 marks
D11v=607, a=2.1, s=900 — find u.
3 marks
Model Answer
607² = u² + 2×2.1×900 368,449 = u² + 3,780 u² = 364,669 u = 603.9 m/s
3 marks
D12v=560, a=1.8, s=1,600 — find u.
3 marks
Model Answer
560² = u² + 2×1.8×1,600 313,600 = u² + 5,760 u² = 307,840 u = 554.8 m/s
