Stopping Distance, Thinking vs Braking Distance, Momentum & Crash Safety
The total stopping distance of a vehicle is the sum of the thinking distance (the distance travelled during the driver's reaction time) and the braking distance (the distance travelled under the braking force). When brakes are applied, work done by friction between the brake pads and discs transfers kinetic energy into thermal energy (½mv² = Fd), which explains why doubling a car's speed quadruples its braking distance. For Higher Tier students, examiners also test the conservation of momentum (p = mv) in collisions and explosions, and how safety features (crumple zones, seatbelts, and airbags) increase impact time to reduce the impact force via F = mΔv / Δt.
- Stopping Distance = Thinking Distance + Braking Distance
- Stopping distance is the total distance a vehicle travels from the moment the driver first spots a hazard to the moment the vehicle comes to a complete stop. Thinking distance is the distance the vehicle travels during the driver's reaction time (between seeing the hazard and applying the brakes). Braking distance is the distance the vehicle travels while slowing down under the braking force.
- Reaction Time (0.2 s to 0.9 s) & Ruler Drop Test
- A typical human reaction time ranges from 0.2 s to 0.9 s. It is increased by tiredness, alcohol, drugs, and distractions (using a phone), and decreased slightly by caffeine. Measured using a computer screen click test or the ruler drop experiment: hold a 30 cm ruler vertically with the 0 cm mark between a partner's thumb and forefinger, drop without warning, record distance d fallen, and calculate reaction time using d = ½gt² -> t = √(2d / 9.8).
- Braking Force, Friction & Energy Transfer (½mv² = Fd)
- When the brake pedal is pushed, brake pads press against the rotating metal wheel discs. Friction between the pads and discs does mechanical work (W = F x d), transferring energy from the kinetic energy store of the car (E_k = ½mv²) to the thermal energy store of the brakes and surroundings (brakes get very hot). Very large decelerations are dangerous because the brakes may overheat (brake fade) or the tyres may lose grip (skidding), and large forces can injure passengers.
Stopping Distance, Thinking vs Braking Distance, Momentum & Crash Safety: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Stopping Distance = Thinking Distance + Braking Distance
- Stopping distance is the total distance a vehicle travels from the moment the driver first spots a hazard to the moment the vehicle comes to a complete stop. Thinking distance is the distance the vehicle travels during the driver's reaction time (between seeing the hazard and applying the brakes). Braking distance is the distance the vehicle travels while slowing down under the braking force.
- Reaction Time (0.2 s to 0.9 s) & Ruler Drop Test
- A typical human reaction time ranges from 0.2 s to 0.9 s. It is increased by tiredness, alcohol, drugs, and distractions (using a phone), and decreased slightly by caffeine. Measured using a computer screen click test or the ruler drop experiment: hold a 30 cm ruler vertically with the 0 cm mark between a partner's thumb and forefinger, drop without warning, record distance d fallen, and calculate reaction time using d = ½gt² -> t = √(2d / 9.8).
- Braking Force, Friction & Energy Transfer (½mv² = Fd)
- When the brake pedal is pushed, brake pads press against the rotating metal wheel discs. Friction between the pads and discs does mechanical work (W = F x d), transferring energy from the kinetic energy store of the car (E_k = ½mv²) to the thermal energy store of the brakes and surroundings (brakes get very hot). Very large decelerations are dangerous because the brakes may overheat (brake fade) or the tyres may lose grip (skidding), and large forces can injure passengers.
- Momentum & Conservation of Momentum (Higher Tier)
- Momentum is a vector quantity defined by: Momentum, p (kg m/s) = Mass, m (kg) x Velocity, v (m/s). Principle of Conservation of Momentum: In a closed system (where no external forces act), the total momentum before an event (a collision or an explosion) is equal to the total momentum after the event.
- Collisions vs Explosions Calculations (Higher Tier)
- 1. Coalescing collision (two objects stick together): m₁u₁ + m₂u₂ = (m₁ + m₂)v. Remember to make velocity negative if an object is initially moving in the opposite direction. 2. Explosion / Recoil (two objects start at rest, so total initial momentum = 0): 0 = m₁v₁ + m₂v₂ -> m₁v₁ = -m₂v₂ (the lighter object moves off at a higher speed in the opposite direction to the heavier object).
- Force as Rate of Change of Momentum (Triple Higher Tier)
- When a non-zero resultant force acts on an object, it causes a change in momentum. Formula: Force, F (N) = Change in momentum, mΔv (kg m/s) / Change in time, Δt (s). Vehicle safety features (crumple zones at front/rear, stretchy seatbelts, inflatable airbags), cycle helmets, and gymnasium crash mats all work by increasing the collision time (Δt) taken for the momentum to reach zero, thereby reducing the rate of change of momentum and reducing the impact force (F).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the equation linking stopping distance, thinking distance, and braking distance.
Does icy weather or worn tyres affect a driver's thinking distance?
At 20 m/s, a car has a thinking distance of 12 m and a braking distance of 24 m. What will its thinking distance and braking distance be at 40 m/s (assuming the same driver and road)?
State the Principle of Conservation of Momentum.
Write the equation linking force, change in momentum, and time (Triple Higher Tier).
Explain in 3 mark-scheme steps how a cycle helmet protects a cyclist's head during a fall (Triple).
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A car of mass 1,200 kg is travelling at 20 m/s when the driver sees a hazard ahead. The driver's reaction time is 0.65 s, and once the brakes are applied, the brakes exert a constant braking force of 6,000 N until the car stops. Calculate: (a) the thinking distance, (b) the braking distance, and (c) the total stopping distance. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) Thinking distance: s_thinking = speed x reaction time = 20 m/s x 0.65 s = 13 m [1 mark]. 2. (b) Initial kinetic energy of the car: E_k = ½ x m x v² = 0.5 x 1,200 x (20)² = 240,000 J [1 mark]. 3. Work done by the brakes equals kinetic energy transferred: W = F x d_braking = 240,000 J [1 mark]. 4. Braking distance: d_braking = 240,000 J / 6,000 N = 40 m (or via a = -F/m = -5 m/s² and v² - u² = 2as) [1 mark]. 5. (c) Total stopping distance = Thinking distance + Braking distance = 13 m + 40 m = 53 m [1 mark].
- Q2
A railway truck A of mass 4,000 kg moving at 5.0 m/s to the right collides with a smaller truck B of mass 2,000 kg moving at 2.0 m/s in the opposite direction (to the left). After the collision, the two trucks lock together and move off at a common velocity v. Calculate the magnitude and direction of velocity v. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Define moving to the right as positive (+) and moving to the left as negative (-). Initial momentum of truck A = 4,000 x (+5.0) = +20,000 kg m/s; Initial momentum of truck B = 2,000 x (-2.0) = -4,000 kg m/s [1 mark]. 2. Total momentum before collision = +20,000 + (-4,000) = +16,000 kg m/s [1 mark]. 3. By conservation of momentum, total momentum after collision = +16,000 kg m/s = (4,000 + 2,000) x v = 6,000v [1 mark]. 4. v = +16,000 / 6,000 = +2.67 m/s (2.67 m/s to the right) [1 mark].
- Q3
A cannon of mass 800 kg is initially stationary. It fires a cannonball of mass 10 kg horizontally at a velocity of +120 m/s. Calculate the recoil velocity of the cannon immediately after firing and explain why the cannon moves backwards. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Before firing, both the cannon and cannonball are stationary, so the total initial momentum of the closed system is 0 kg m/s [1 mark]. 2. By the Principle of Conservation of Momentum, the total momentum after firing must also equal 0 kg m/s [1 mark]. 3. Forward momentum of cannonball = 10 kg x (+120 m/s) = +1,200 kg m/s; therefore, momentum of cannon = -1,200 kg m/s [1 mark]. 4. Recoil velocity of cannon: v = -1,200 / 800 = -1.5 m/s (1.5 m/s backwards, in the opposite direction to the cannonball) [1 mark].
- Q4
During a crash test, a car travelling at 15 m/s hits a concrete wall. A crash-test dummy of mass 70 kg is brought to rest in 0.12 s by a seatbelt and airbag. Calculate the average deceleration force acting on the dummy, and explain how the seatbelt and airbag reduce the risk of serious injury compared to hitting the rigid dashboard in 0.02 s. (Triple Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Calculate the change in momentum of the dummy: mΔv = 70 kg x (0 - 15 m/s) = -1,050 kg m/s [1 mark]. 2. Using F = mΔv / Δt: Average force with seatbelt/airbag = -1,050 / 0.12 = -8,750 N (magnitude 8,750 N) [1 mark]. 3. The seatbelt stretches and the airbag compresses when the dummy hits them, which increases the time taken (Δt) for the dummy's momentum to fall to zero (from 0.02 s to 0.12 s, a 6-fold increase) [1 mark]. 4. Because F = mΔv / Δt, increasing the collision time for the same change in momentum reduces the rate of change of momentum, reducing the impact force on the dummy by 6 times (from 52,500 N down to 8,750 N) [1 mark].
Common Exam Questions (4)
Which factors affect only thinking distance, which affect only braking distance, and which factor affects both?
1. Speed is the only factor that affects both thinking distance and braking distance. 2. Thinking distance only is increased by factors that increase driver reaction time: tiredness, alcohol, drugs, and distractions (e.g., mobile phones). 3. Braking distance only is increased by poor road/weather conditions (wet, icy, or oily roads) and poor vehicle condition (worn tyres with little tread or worn/faulty brakes), as well as a heavier vehicle mass.
Why does doubling the speed of a car double the thinking distance, but QUADRUPLE (x4) the braking distance?
Thinking distance = speed x reaction time (s = vt); since reaction time is constant, doubling v doubles s. For braking distance, the work done by the braking force equals the initial kinetic energy of the car: F x d = ½ x m x v². Because kinetic energy is proportional to v² (speed squared), doubling the speed (2v) increases the kinetic energy by a factor of 2² = 4, so for the same maximum braking force (F), the braking distance (d) must be 4 times longer.
How do seatbelts, airbags, and crumple zones reduce injury in a car crash? (Triple Higher Tier)
Use the rate of change of momentum equation: Force = Change in momentum / Time taken (F = mΔv / Δt). In a crash, the car and passengers must lose all their momentum (a fixed change in momentum, mΔv, from initial speed to zero). Crumple zones, seatbelts, and airbags deform/stretch, which increases the distance and increases the time taken (Δt) for the passenger to come to rest. Dividing by a larger time (Δt) reduces the rate of change of momentum, which decreases the impact force (F) acting on the passenger.
Why must you assign positive and negative signs to velocities in momentum calculations? (Higher Tier)
Momentum (p = m x v) is a vector quantity, so direction matters. Always define moving to the right (or forwards) as positive (+) and moving to the left (or backwards/rebounding) as negative (-). In an explosion (such as firing a cannon or two ice skaters pushing apart from rest), the total initial momentum is 0 kg m/s, so the total final momentum must also sum to 0 kg m/s (meaning one momentum is +mv and the other is -mv).
Need more questions on Stopping Distance, Thinking vs Braking Distance, Momentum & Crash Safety, or want to organise your own notes?
Type a specific exam question you are stuck on, pick one of the prompts below, or paste your class notes to build a custom set of flashcards and practice questions on this page.
Related study guides
Forces, Motion & Momentum
A force is a push or pull that acts on an object due to its interaction with another object, capable of changing its speed, direction, or shape. This guide covers scalar vs vector quantities, resultant forces, Newton's three laws of motion, terminal velocity, stopping distances, and conservation of momentum.
Newton's Laws of Motion
Published in 1687, Newton's three laws of motion form the foundation of classical mechanics across GCSE and A-Level Physics. They relate the forces acting on a body to its state of motion: equilibrium when resultant force is zero (First Law), acceleration proportional to resultant force via F = ma (Second Law), and equal and opposite interaction pairs between two bodies (Third Law).
Distance-Time & Velocity-Time Graphs, Acceleration & Terminal Velocity
Motion in a straight line is analysed using vector quantities (displacement, velocity, and acceleration) that have both magnitude and direction, alongside scalar quantities (distance and speed) that have magnitude only. On a distance-time graph, the gradient represents speed, whereas on a velocity-time graph, the gradient represents acceleration (a = Δv / t) and the area enclosed under the graph represents the total distance travelled (displacement). Examiners frequently test multi-stage velocity-time graph area calculations, the uniform acceleration equation v² - u² = 2as, and the 6-mark explanation of terminal velocity using Newton's First and Second Laws.