NSWEngineering StudiesSyllabus dot point

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 OpusReviewed by Better Tuition Academy6 min answerUpdated 2026-05-18

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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.

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.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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.