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Engineering mechanics: How are Newton's laws used to analyse acceleration, braking and crash performance of vehicles?

Apply Newton's laws of motion to road vehicles, calculate accelerating and braking forces, and analyse impulse and momentum in crashes

A focused answer to the HSC Engineering Studies Personal and Public Transport dot point on Newton's laws. Acceleration and braking force on a vehicle, impulse and momentum in collisions, the crumple zone, ANCAP testing, and worked HSC-style past exam questions.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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In plain English

Think about catching a fast cricket ball. If you snatch it still with stiff, straight arms, it stings, because you stop the ball's momentum in a fraction of a second and the force on your hands is huge. If instead you let your arms "give" and move backward with the ball for half a second before it stops, catching the exact same ball at the exact same speed barely hurts at all. Nothing about the ball changed, only how long it took to stop. A car crash works exactly the same way: the change in momentum ( $mass \times change\ in\ speed$ ) is fixed by how fast the car was going and how heavy it is, so the only lever engineers have is time. A crumple zone is the "giving arms" of the car, deliberately built to buckle and extend the stopping time from a violent instant to a survivable fraction of a second, which is why $F \Delta t = \Delta p$ (force times time equals change in momentum) is the single most useful equation in vehicle safety engineering.

What this dot point is asking

NESA wants you to apply Newton's three laws of motion to road vehicles: calculate traction and braking forces, find acceleration from a known engine or brake force, and analyse collisions using impulse and momentum.

The answer

Newton's first law in vehicles

A vehicle continues in uniform motion unless acted on by an unbalanced force. The forces on a moving car are:

Tractive effort from the driven wheels (engine torque divided by wheel radius and reduced by drivetrain efficiency)
Aerodynamic drag $F_d = \frac{1}{2} \rho v^2 C_d A$
Rolling resistance $F_r = \mu_r m g$ (about 1.5 percent of weight for road tyres)
Gravity along the road grade (relevant on hills)

At constant cruise speed, tractive effort equals the sum of drag and rolling resistance.

An owned free-body diagram makes the four forces on a cruising vehicle concrete: traction pushes the car forward, drag and rolling resistance push it backward, and weight and the road's normal force balance vertically.

Newton's second law

F = ma

The net force determines the acceleration. A 1500 kg sedan accelerating at $3 \text{ m/s}^2$ requires net forward force $F = 1500 \times 3 = 4500$ N.

For braking, the friction force from the brake pads on the discs decelerates the wheels, and the friction between tyre and road decelerates the vehicle. Maximum deceleration is limited by tyre-road friction:

a_{\max} = \mu g

For dry road, $\mu \approx 0.8$ , giving $a_{\max} \approx 7.8 \text{ m/s}^2$ . For wet road, $\mu \approx 0.4$ , giving $a_{\max} \approx 3.9 \text{ m/s}^2$ .

Newton's third law

Every action force has an equal and opposite reaction. The drive tyre pushes the road backward; the road pushes the tyre forward by the same force. This is the source of propulsion on land vehicles.

Impulse and momentum in collisions

F \Delta t = \Delta p = m \Delta v

For a given change in momentum (which is fixed by the impact speed and vehicle mass), extending the stopping time reduces the average force. Crumple zones, airbags and seatbelt webbing all extend $\Delta t$ during impact, reducing the peak force on occupants.

The ANCAP (Australasian New Car Assessment Program) tests cars at 50 km/h frontal offset, 60 km/h side impact and 75 km/h oblique pole impact and scores body shell deformation, dummy chest and head decelerations, and post-crash fire risk. ANCAP star ratings drive Australian vehicle design and purchasing decisions.

An owned force-time diagram shows why the same crash produces two very different peak forces: the area under each pulse (the impulse) is fixed by the impact momentum, so a wider pulse must be shorter.

Energy in braking and stopping distance

Braking converts the vehicle's kinetic energy into heat at the brakes. The work-energy theorem gives the braking distance from a constant deceleration:

\tfrac{1}{2} m v^2 = F_b \, s \qquad\Rightarrow\qquad s = \frac{v^2}{2a}

Because stopping distance depends on the square of speed, doubling the speed quadruples the braking distance. Total stopping distance also includes the reaction distance ( $v \times t_{\text{reaction}}$ , typically 1.5 s of travel before braking begins), which is why posted speed limits have such a large effect on crash outcomes. The brakes must dissipate this energy as heat, so disc size and ventilation are sized for repeated high-energy stops (for example, a bus descending a long grade).

Friction, traction and the role of ABS

All tractive and braking forces are transmitted through the tyre contact patches and are limited by friction, $F_{\max} = \mu N$ . The static (rolling) friction coefficient is higher than the kinetic (sliding) value, so a tyre that is rotating without slipping grips better than one that is locked and skidding. Anti-lock braking systems (ABS) rapidly modulate brake pressure to keep the wheels just below lock-up, holding the tyre in the higher static-friction regime, which shortens stopping distance and preserves steering control. Electronic stability control extends the same principle to cornering.

Marker's note: two model answers

Example A: braking force and distance. A 1200 kg car brakes from 100 km/h (27.8 m/s) to rest in 4 s.

Step 1: find the average deceleration. $a = \Delta v / \Delta t = 27.8 / 4 = 6.94 \text{ m/s}^2$ .

Step 2: find the average braking force. $F = ma = 1200 \times 6.94 = 8330$ N.

Step 3: find the stopping distance. $s = v t / 2 = 27.8 \times 4 / 2 = 55.6$ m.

Marker's note: full marks need all three quantities in order (deceleration, then force, then distance), each with correct units, because later steps depend on earlier ones; a bare final number with no working loses the method marks even if it happens to be correct.

Example B: impulse and crash force. The same 1200 kg car instead strikes a barrier at 15 m/s and its passenger cell decelerates to rest in 0.08 s because the crumple zone deforms.

Step 1: find the change in momentum. $\Delta p = m \Delta v = 1200 \times 15 = 18{,}000$ kg m/s.

Step 2: find the average force. $F = \Delta p / \Delta t = 18{,}000 / 0.08 = 2.25 \times 10^5 \text{ N} = 225$ kN.

Step 3: compare with no crumple zone. If the same car stopped in 0.02 s instead, $F = 18{,}000 / 0.02 = 9.0 \times 10^5 \text{ N} = 900$ kN, four times larger for the identical impact.

Marker's note: markers reward stating $F \Delta t = \Delta p$ explicitly before substituting, carrying units through every line, and, where a comparison is asked for, quoting the ratio (here, fourfold) rather than just leaving two separate numbers for the reader to compare.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC style5 marksA 1500 kg vehicle travelling at 60 km/h collides with a rigid barrier. Without crumple zones, the vehicle stops in 0.05 s. With crumple zones, it stops in 0.2 s. Calculate the average deceleration force in each case and explain how crumple zones reduce occupant injury.

Show worked answer →

Convert the speed. $60 \text{ km/h} = 60 / 3.6 = 16.67 \text{ m/s}$ .

Change in momentum: $\Delta p = m \Delta v = 1500 \times 16.67 = 25{,}000$ kg m/s.

Without crumple zones ( $\Delta t = 0.05$ s):

F = \frac{\Delta p}{\Delta t} = \frac{25{,}000}{0.05} = 5.0 \times 10^5 \text{ N} = 500 \text{ kN}

With crumple zones ( $\Delta t = 0.2$ s):

F = \frac{25{,}000}{0.2} = 1.25 \times 10^5 \text{ N} = 125 \text{ kN}

Crumple zones extend the stopping time by a factor of four, which divides the average force on the occupants by the same factor. The same change in momentum is spread over a longer interval, so the rate of change of momentum (the force) is lower.

Engineering implication: the front and rear of the body shell are designed to deform progressively at controlled load levels, absorbing kinetic energy through plastic deformation of high-strength low-alloy steel pressings. The passenger cell is built from ultra-high-strength boron steel and stays largely undeformed to preserve survival space.

Markers reward (1) the impulse-momentum equation $F \Delta t = \Delta p$ , (2) both numerical answers with units, (3) the engineering link between time extension and force reduction, and (4) reference to crumple zone deformation absorbing energy.

HSC 20212 marksUsing Newton's third law, explain how a road vehicle is propelled forward.

Show worked answer →

The engine torque turns the driven tyres, which push backward against the road surface (the action force). By Newton's third law the road pushes forward on the tyres with an equal and opposite reaction force, and this forward reaction is the traction that propels the vehicle. Markers award one mark for identifying the action-reaction pair and one for stating that the road's forward reaction on the tyre is the driving force.

HSC 20246 marksA 1400 kg car accelerates from rest. The driven wheels can transmit a maximum tractive force limited by tyre-road friction with a coefficient of 0.8. Aerodynamic drag and rolling resistance total 600 N at low speed. Determine the maximum possible acceleration from rest, the time to reach 100 km/h at this acceleration, and comment on one factor that would reduce the acceleration achieved in practice.

Show worked answer →

Maximum tractive force (friction limit).

F_{\text{trac}} = \mu m g = 0.8 \times 1400 \times 9.81 = 1.099 \times 10^4 \text{ N} = 10.99 \text{ kN}

Net accelerating force.

F_{\text{net}} = F_{\text{trac}} - F_{\text{resist}} = 10{,}990 - 600 = 10{,}390 \text{ N}

Maximum acceleration.

a = \frac{F_{\text{net}}}{m} = \frac{10{,}390}{1400} = 7.42 \text{ m/s}^2

Time to 100 km/h. $100 \text{ km/h} = 27.8 \text{ m/s}$ , so

t = \frac{v}{a} = \frac{27.8}{7.42} = 3.75 \text{ s}

Practical comment. Real acceleration is lower because tractive force is limited by available engine torque at low speeds (and by traction control), drag and rolling resistance rise with speed, and not all weight sits on the driven axle. Markers reward the friction-limited tractive force, the net force and acceleration, the time from $v = at$ , and a sensible limiting factor.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksA 1800 kg delivery van accelerates uniformly from rest to 12 m/s in 6 s on a flat road. Calculate the net accelerating force required, ignoring resistive forces.

Show worked solution →

Step 1: find acceleration.

a = \frac{\Delta v}{\Delta t} = \frac{12}{6} = 2.0 \text{ m/s}^2

Step 2: apply Newton's second law.

F = ma = 1800 \times 2.0 = 3600 \text{ N}

Marking criteria: 1 mark for correct acceleration, 1 mark for correctly applying $F = ma$ , 1 mark for the correct final answer with unit (3600 N).

foundation3 marksExplain, using Newton's third law, why a car cannot accelerate forward on a sheet of frictionless ice, even with the engine running at full power.

Show worked solution →

Newton's third law requires the driven tyre to push backward on the road surface (the action force) so that the road pushes forward on the tyre with an equal and opposite reaction force, which is the traction that propels the car.

On frictionless ice there is no friction between the tyre and the surface, so the tyre cannot exert a backward action force on the ice; it simply spins in place. With no action force, there is no reaction force, so no traction force acts on the car and it cannot accelerate forward regardless of engine power.

Marking criteria: 1 mark for identifying the tyre-road action-reaction pair, 1 mark for stating that friction is required to transmit the action force, 1 mark for the correct conclusion that zero friction means zero traction.

core5 marksA road-safety test compares two identical 1300 kg cars braking from 20 m/s (72 km/h) to rest on different surfaces. Car A brakes on dry bitumen with tyre-road friction coefficient 0.80. Car B brakes on a wet, poorly drained section of the same road with friction coefficient 0.35. (a) Calculate the maximum deceleration available to each car. (b) Calculate the minimum braking distance for each car. (c) State, with a reason, which surface condition NESA-style road-safety campaigns most often target when reducing speed limits.

Show worked solution →

(a) Maximum deceleration, $a_{\max} = \mu g$ .

a_A = 0.80 \times 9.81 = 7.85 \text{ m/s}^2 \qquad a_B = 0.35 \times 9.81 = 3.43 \text{ m/s}^2

(b) Minimum braking distance, $s = v^2 / (2a)$ .

s_A = \frac{20^2}{2 \times 7.85} = 25.5 \text{ m} \qquad s_B = \frac{20^2}{2 \times 3.43} = 58.3 \text{ m}

(c) Wet or poorly drained roads. Car B needs more than double the braking distance of Car A for the identical starting speed and mass, purely because the available friction coefficient is lower. Campaigns target wet-weather speed limits and road drainage because the same driver behaviour produces a much longer, less predictable stopping distance once friction drops.

Marking criteria: 1 mark each for the two decelerations, 1 mark each for the two braking distances, 1 mark for a reasoned comparison linking the near tripling of stopping distance to the lower friction coefficient (not just restating the numbers).

core5 marksA 1250 kg hatchback travelling at 18 m/s strikes a parked truck. Onboard sensors show the passenger cell decelerates to rest over 0.15 s once the airbags and crumple zone are fully engaged. Calculate (a) the change in momentum of the car, (b) the average force on the passenger cell during the crash, and (c) the average force if the same car had no crumple zone and stopped in 0.03 s instead.

Show worked solution →

(a) Change in momentum.

\Delta p = m \Delta v = 1250 \times 18 = 22{,}500 \text{ kg m/s}

(b) Average force with crumple zone ( $\Delta t = 0.15$ s).

F = \frac{\Delta p}{\Delta t} = \frac{22{,}500}{0.15} = 1.5 \times 10^5 \text{ N} = 150 \text{ kN}

(c) Average force without crumple zone ( $\Delta t = 0.03$ s).

F = \frac{22{,}500}{0.03} = 7.5 \times 10^5 \text{ N} = 750 \text{ kN}

Removing the crumple zone increases the average force fivefold for the identical crash, because the same momentum change now occurs five times faster.

Marking criteria: 1 mark for correct $\Delta p$ , 1 mark for correct force with crumple zone, 1 mark for correct force without crumple zone, 1 mark for both answers carrying correct units, 1 mark for a stated comparison (fivefold increase) linking the shorter time to the larger force.

core4 marksUsing the data below from a simplified ANCAP-style frontal offset test, calculate the average force experienced by a 75 kg crash-test dummy if its upper body decelerates from the 50 km/h test speed to rest over the stated contact time. Test data: impact speed 50 km/h; dummy chest contact time with the seatbelt and airbag system, 0.11 s.

Show worked solution →

Step 1: convert speed.

50 \text{ km/h} = \frac{50}{3.6} = 13.9 \text{ m/s}

Step 2: change in momentum of the dummy.

\Delta p = m \Delta v = 75 \times 13.9 = 1042 \text{ kg m/s}

Step 3: average force.

F = \frac{\Delta p}{\Delta t} = \frac{1042}{0.11} = 9470 \text{ N} \approx 9.5 \text{ kN}

Marking criteria: 1 mark for correct speed conversion, 1 mark for correct change in momentum, 1 mark for correct average force, 1 mark for a sensible unit and rounding (approximately 9.5 kN).

exam7 marksAssess the effectiveness of crumple zones and airbags, used together, as engineering solutions for reducing occupant injury in a frontal collision. Refer to impulse-momentum theory and to ANCAP testing in your answer.

Show worked solution →

This is a 7-mark ASSESS: markers reward a supported judgement, not just a description of each feature.

Plan.

Thesis: crumple zones and airbags are highly effective specifically because they attack the SAME variable, the collision time $\Delta t$ , at two different stages of the crash, and ANCAP testing confirms this with measured occupant deceleration data rather than assumption.
Physics: for a fixed change in momentum $\Delta p = m \Delta v$ set by the impact speed and vehicle mass, the impulse-momentum relationship $F \Delta t = \Delta p$ shows average force falls as $\Delta t$ rises. Neither feature changes the total energy that must be dissipated; both change the rate at which the associated force acts.
Crumple zone: the vehicle's body shell deforms progressively (plastic deformation of high-strength steel pressings), extending the time over which the whole vehicle decelerates from tens of milliseconds to roughly 100 to 150 ms, while the boron-steel passenger cell stays largely rigid to preserve survival space.
Airbag: acts on the shorter, later timescale, over the final tens of milliseconds as the occupant's body decelerates relative to the now-stationary cell, cushioning the head and chest so their individual $\Delta t$ is extended further than a seatbelt alone would allow.
Evidence: ANCAP frontal offset testing at 50 km/h measures instrumented dummy chest and head decelerations directly; cars combining strong crumple-zone structures with airbags consistently score higher adult-occupant protection ratings than cars with either feature alone.
Limitation/judgement: neither feature helps if the impact speed is very high (structure "bottoms out" before the passenger cell), and airbags can injure very small or out-of-position occupants; overall, though, for typical frontal-offset speeds the two features are effective because they lower peak force at two sequential stages of the same collision, and this is independently confirmed by measured ANCAP dummy data rather than theory alone.

Marker's note: top-band responses (1) apply $F \Delta t = \Delta p$ explicitly and correctly, (2) explain the two features as acting on the SAME physical quantity ( $\Delta t$ ) at different, sequential stages of the crash rather than describing them independently, (3) cite ANCAP as the evidence base, and (4) close with an explicit, qualified judgement rather than a neutral summary.

exam6 marksA 1600 kg all-wheel-drive car can use all four tyres for traction, with a tyre-road friction coefficient of 0.90 on dry bitumen. Combined aerodynamic drag and rolling resistance at low speed total 550 N. (a) Calculate the maximum possible accelerating force and the resulting maximum acceleration from rest. (b) A front-wheel-drive version of the same car, with the same total mass, can only use the front axle (55 percent of the vehicle's weight) for traction. Calculate its maximum acceleration and explain the difference between the two drivetrains.

Show worked solution →

(a) All-wheel-drive.

F_{\text{trac}} = \mu m g = 0.90 \times 1600 \times 9.81 = 1.412 \times 10^4 \text{ N}

F_{\text{net}} = 14{,}120 - 550 = 13{,}570 \text{ N} \qquad a = \frac{F_{\text{net}}}{m} = \frac{13{,}570}{1600} = 8.48 \text{ m/s}^2

(b) Front-wheel-drive. Only 55 percent of the weight sits on the driven (front) axle, so the normal force available for traction is reduced:

F_{\text{trac}} = \mu (0.55 \, mg) = 0.90 \times 0.55 \times 1600 \times 9.81 = 7766 \text{ N}

F_{\text{net}} = 7766 - 550 = 7216 \text{ N} \qquad a = \frac{7216}{1600} = 4.51 \text{ m/s}^2

Explanation. Both cars have identical mass, tyres and friction coefficient, but the all-wheel-drive car can use the full vehicle weight to generate traction while the front-wheel-drive car is limited to the normal force on its driven axle alone. Since $F_{\text{trac}} = \mu N$ , halving (roughly) the usable normal force nearly halves the maximum tractive force and hence the achievable acceleration, even though total friction available at all four contact patches is unchanged.

Marking criteria: 1 mark for correct AWD tractive force, 1 mark for correct AWD acceleration, 1 mark for correctly reducing the normal force to the front-axle share, 1 mark for correct FWD tractive force, 1 mark for correct FWD acceleration, 1 mark for an explanation linking the difference to the normal force term in $F_{\text{trac}} = \mu N$ rather than the friction coefficient (which is unchanged).