🚀 Dynamics — Newton's Laws & Kinematics
Dynamics is the study of objects in motion and the forces that cause or change that motion. Engineers use Newton's laws and the kinematic equations to design vehicle braking systems, predict projectile paths, specify motor drives and size shafts for rotating machinery.
2Topics
8Examples
4Calculators
12Practice Qs
🚀 Level 3 context: Every time you size a drive motor, calculate stopping distance, or analyse a crash, you need Newton's Second Law and the kinematic equations. The four SUVAT equations let you find any one of the five variables when you know the other four.
🚀Newton's LawsF = ma, impulse, momentum
📈KinematicsSUVAT equations
⚙️Applications4 dynamics calculators
🎮Game12 randomised questions
📝Practice12 graded questions
🚀
Newton's Laws & Linear Motion
Three laws that govern every moving object: inertia, F = ma, and action–reaction. Combined with impulse and momentum, they solve force, mass and acceleration problems.
🎮Live explorer — F = maDrag sliders — car motion diagram updates
💡 Double the force → double the acceleration. Double the mass → halve the acceleration.
📋Try it — Newton's 2nd Law CalculatorUpdates as you type
Select what to find and enter values above.
📚 Key vocabulary
- Inertia
- the tendency of an object to resist changes in its state of motion — proportional to mass.
- Resultant force
- the single net force acting on a body after all forces are combined vectorially.
- Momentum (p)
- mass × velocity (kg m/s) — a vector quantity in the direction of motion.
- Impulse
- force × time (N s) — equals the change in momentum: FΔt = Δ(mv).
- Friction force
- F = μN — coefficient of friction × normal reaction. Opposes relative motion.
- Free body diagram
- a diagram showing all forces acting on a single body, used before applying F = ma.
👁️ Visual aid — free fall: a 5 kg mass falls under gravity. Only weight acts downward (g = 9.81 m/s²)
✍️Worked example 1 — force to accelerate a vehicle step through it
QuestionA 1500 kg car accelerates from rest at 2.4 m/s². Calculate the resultant driving force needed.
Formula
F = m × aSubstitute
F = 1500 × 2.4Multiply
1500 × 2.4 = 3600AnswerF = 3600 N ✓
Step 1 of 5
✍️Worked example 2 — deceleration with friction
QuestionA 900 kg machine slides to rest from 6 m/s. Friction coefficient μ = 0.4. Find friction force and deceleration.
Normal reaction
N = mg = 900 × 9.81 = 8829 NFriction force
F = μN = 0.4 × 8829 = 3531.6 NDeceleration
a = F ÷ m = 3531.6 ÷ 900 = 3.924 m/s²AnswerF = 3531.6 N a = 3.924 m/s² (deceleration)
✍️Worked example 3 — impulse and change in momentum
QuestionA 0.5 kg ball hits a wall at 12 m/s and rebounds at 8 m/s. The impact lasts 0.02 s. Find the impulse and average force.
Change in momentum
Δp = m(v − u) = 0.5 × (8 − (−12)) = 0.5 × 20 = 10 N sAverage force
F = Δp ÷ t = 10 ÷ 0.02 = 500AnswerImpulse = 10 N s Average force = 500 N
✍️Worked example 4 — weight and mass on an inclined plane
QuestionA 200 kg crate rests on a frictionless ramp at 25°. Find the force needed to hold it stationary.
Weight
W = 200 × 9.81 = 1962 NComponent down ramp
F = W sin 25° = 1962 × 0.4226 = 829.1AnswerF = 829.1 N (up the ramp to hold it)
📈
Kinematics — SUVAT Equations
The four equations of uniform motion link displacement, initial and final velocity, acceleration and time. Know three quantities and you can find the other two.
🎮Live explorer — SUVAT: find v and sDrag sliders to explore
Final velocity v
15.0 m/s
v = u + at
Displacement s
37.5 m
s = ut + ½at²
v² check
225
v² = u² + 2as
💡 All three equations give consistent answers — pick the one that avoids the unknown you don't need.
📋Try it — SUVAT CalculatorChoose the equation — updates as you type
Select an equation and enter the known values.
📚 Key vocabulary
- Displacement (s)
- change in position — a vector with direction. Not the same as distance if the object changes direction.
- Velocity (v, u)
- rate of change of displacement — speed with direction. Units: m/s.
- Acceleration (a)
- rate of change of velocity. Positive = speeding up (same direction); negative = slowing down.
- Uniform acceleration
- constant acceleration — the SUVAT equations apply only when acceleration is constant.
- Free fall
- motion under gravity alone: a = g = 9.81 m/s² downward (or take upward as positive, a = −9.81).
- SUVAT
- mnemonic for the five kinematic variables: s, u, v, a, t. Know three, find the other two.
👁️ Visual aid — vehicle accelerating from u = 0, a = 3 m/s². Watch v and s grow over time.
✍️Worked example 5 — stopping distance (v² = u² + 2as) step through it
QuestionA car brakes from 20 m/s to rest with a deceleration of 5 m/s². Find the stopping distance.
Known
u = 20 m/s, v = 0, a = −5 m/s² (deceleration)Equation
v² = u² + 2as → rearrange: s = (v² − u²) ÷ (2a)Substitute
s = (0 − 400) ÷ (2 × (−5)) = −400 ÷ −10Calculate
s = 400 ÷ 10 = 40AnswerStopping distance = 40 m ✓
Step 1 of 6
✍️Worked example 6 — time to reach a velocity
QuestionA machine part accelerates from rest at 2.5 m/s². How long to reach 15 m/s?
Equation
v = u + at → t = (v − u) ÷ aSubstitute
t = (15 − 0) ÷ 2.5 = 15 ÷ 2.5Answert = 6 s
✍️Worked example 7 — displacement using s = ut + ½at²
QuestionA vehicle starts at 5 m/s and accelerates at 1.5 m/s². Find its displacement after 8 s.
Equation
s = ut + ½at²Substitute
s = 5×8 + ½×1.5×64 = 40 + 48Answers = 88 m
✍️Worked example 8 — acceleration from v² = u² + 2as
QuestionA ball decelerates from 18 m/s to 6 m/s over 24 m. Find the deceleration.
Equation
a = (v² − u²) ÷ (2s)Substitute
a = (36 − 324) ÷ (2 × 24) = −288 ÷ 48Answera = −6 m/s² (deceleration of 6 m/s²)
⚙️
Engineering Applications
Four dynamics calculators for vehicle, machine and impact scenarios.
🚘 Vehicle braking — stopping distance
Find the stopping distance for any vehicle from its initial speed and braking deceleration.
s = (v² − u²) ÷ 2a (v = 0 at rest)
🧮 Calculator
Enter initial speed and deceleration.
🚀 Drive motor — force and acceleration
Find the resultant force and acceleration when a motor drives a known mass.
a = F ÷ m (or rearrange for F or m)
🧮 Calculator
Enter drive force and mass.
📷 Impact — impulse and average force
Calculate the impulse and average impact force from mass, velocity change and contact time.
Impulse = mΔv F = mΔv ÷ t
🧮 Calculator
Enter mass, velocity change and contact time.
📈 SUVAT — find any variable
Given initial velocity, acceleration and time, find final velocity and displacement in one step.
v = u + at s = ut + ½at²
🧮 Calculator
Enter u, a and t to find v and s.
🎮
Dynamics Challenge
12 randomised questions — Newton's laws and SUVAT.
🚀 Quick-fire round
Question 1 / 12
Score: 0
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Press Start to begin!
📝
Practice Questions
12 graded questions — type your answer to check, then reveal full working.
🚀 State which SUVAT equation or Newton's law you are using. Show substitution and every step. g = 9.81 m/s².
Q1 · F = ma1 mark
A 2000 kg vehicle accelerates at 1.8 m/s². Calculate the resultant driving force.
F = ma = 2000 × 1.8 = 3600 N
Q2 · F = ma2 marks
A resultant force of 5400 N acts on a 1350 kg mass. Calculate the acceleration.
a = F ÷ m = 5400 ÷ 1350 = 4 m/s²
Q3 · Weight1 mark
A machine has mass 350 kg. Calculate its weight (g = 9.81 m/s²).
W = mg = 350 × 9.81 = 3433.5 N
Q4 · Impulse2 marks
A 0.8 kg ball's velocity changes from 10 m/s to −6 m/s on impact. Find the impulse.
Δp = m(v−u) = 0.8×(−6−10) = 0.8×(−16) = −12.8 N s (magnitude = 12.8 N s)
Q5 · v = u + at1 mark
A vehicle starts at 8 m/s and accelerates at 2.5 m/s² for 6 s. Find the final velocity.
v = u + at = 8 + 2.5×6 = 8 + 15 = 23 m/s
Q6 · s = ut + ½at²2 marks
A part starts from rest and accelerates at 4 m/s² for 5 s. Find the displacement.
s = ut + ½at² = 0 + ½×4×25 = 50 m
Q7 · v² = u² + 2as2 marks
A vehicle brakes from 25 m/s to rest over 62.5 m. Find the deceleration.
a = (v²−u²)÷(2s) = (0−625)÷(125) = −5 m/s² (deceleration = 5 m/s²)
Q8 · Find time2 marks
A machine accelerates from 4 m/s to 16 m/s at 3 m/s². Find the time taken.
t = (v−u)÷a = (16−4)÷3 = 12÷3 = 4 s
Q9 · Friction2 marks
A 600 kg crate slides on a floor with μ = 0.35. Find the friction force and deceleration.
N = 600×9.81 = 5886 N
F = μN = 0.35×5886 = 2060.1 N
a = F÷m = 2060.1÷600 = 3.43 m/s²
F = μN = 0.35×5886 = 2060.1 N
a = F÷m = 2060.1÷600 = 3.43 m/s²
Q10 · Inclined plane3 marks
A 300 kg trolley is pushed up a frictionless 20° ramp. Find the force needed to keep it moving at constant speed. (sin 20° = 0.342)
W = 300×9.81 = 2943 N
F = W sinθ = 2943×0.342 = 1006.5 N
F = W sinθ = 2943×0.342 = 1006.5 N
Q11 · Combined3 marks
A 1200 kg car applies 4800 N drive force against 800 N resistance. Find (a) resultant force, (b) acceleration, (c) velocity after 10 s from rest.
R = 4800−800 = 4000 N
a = 4000÷1200 = 3.33 m/s²
v = 0+3.33×10 = 33.3 m/s
a = 4000÷1200 = 3.33 m/s²
v = 0+3.33×10 = 33.3 m/s
Q12 · Combined3 marks
A ball is thrown upward at 15 m/s. Find (a) time to reach maximum height, (b) maximum height, (c) velocity after 2 s. (g = 9.81 m/s²)
t = 15÷9.81 = 1.53 s to max height
h = (15²)÷(2×9.81) = 225÷19.62 = 11.47 m
v at 2s = 15−9.81×2 = 15−19.62 = −4.62 m/s (falling)
h = (15²)÷(2×9.81) = 225÷19.62 = 11.47 m
v at 2s = 15−9.81×2 = 15−19.62 = −4.62 m/s (falling)