071 Physics of Vehicles Safety.pdf

Physics Factsheet

www.curriculum-press.co.uk

Number 71

Physics of Vehicles: Safety This shows us that the braking distance is proportional to the square of the speed. The minus sign just appears because this is deceleration - so a has a negative sign

This Factsheet focuses on applying concepts from mechanics to motor vehicles. It covers: • stopping distances • car safety features • analysing accidents Factsheet 66 deals with how cars move and brake, and using the ideas of motive force and power.

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Before working through this Factsheet you should understand • the relationship between force, work and energy • Newton's second law • how to use the equations of motion (Factsheet 13)

Thinking distance is proportional to the speed Braking distance is proportional to the square of the speed

The typical value for deceleration used for most cars in good conditions is 6.58 ms-2 (you do not have to remember this!). But this will vary depending on the weather and the condition of the car - braking distances will • double in wet weather

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increase by a factor of ten if it is icy This is because of the reduced friction in these conditions. Similarly, if tyre treads are in poor condition, reduced friction again results in increased breaking distances

Stopping Distances Knowing the distance in which a car can stop is vitally important for safe driving. It helps you decide: • how early you must start braking before a traffic light

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The table below shows how thinking, braking and stopping distances vary with speed.

the minimum safe distance to allow between you and the next car the fastest you can safely drive when you can't see far ahead

Note: the stopping distances only give an indication - in practice, drivers should allow a margin for error - i.e. drive slower or allow a greater distance.

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There are two components to stopping distances: "Thinking distance" - the distance the car travels while the driver is reacting (i.e. between them seeing the hazard and applying the brakes) "Braking distance" - the distance the car travels from when the brakes are applied until it comes to rest.

dT /m

4.00 8.00 12.00 16.00 20.00 24.00 28.00 32.00 36.00

2.67 5.33 8.00 10.67 13.33 16.00 18.67 21.33 24.00

dB/m

dS/m

1.21 4.86 10.93 19.44 30.37 43.74 59.53 77.75 98.41

v = speed dB = braking distance

3.88 10.19 18.93 30.10 43.71 59.74 78.20 99.09 122.41

dT = thinking distance dS = stopping distance

The largest contribution to overall stopping distance - particularly at large distances - is from the braking distance (as shown in the graph below). This demonstrates the importance of ensuring the car is kept in optimum condition - however quick your reactions are, they cannot make up for inefficient braking!

Thinking distance This is determined just by the speed of the driver's reactions. For most people, this is about two thirds of a second (although tiredness, or alcohol or drug use, will increase this substantially). While the driver is reacting, the car will still be travelling at its original (constant) speed.

120

Thinking distance is given by: Thinking distance = speed × reaction time Using a "typical" reaction time, this gives: dT = 2/3v

100 80 distance/m

Braking distance This is determined just by the deceleration of the car - which in turn is determined by the braking force applied to the car.

v/ms-1

If we assume this deceleration is constant, then using the equation 2as = v2 − u2 -u2 with v (the final velocity) as zero, we obtain: s = 2a

total stopping distance

60

braking distance

40 20 thinking distance 0 0

1

5

10 15 speed/ms-1

20

25

30

35

Physics Factsheet

71 - Physics of vehicles: safety

Activity: stopping distances in the Highway Code In the UK, speeds of cars are quoted in miles per hour rather than metres per second. You can set up a spreadsheet to calculate thinking, braking and stopping distances from speeds in miles per hour as follows: 1. Enter your speeds in miles per hour into the first column 2. Set up the second column as speeds in metres per second. To convert miles per hour to metres per second, the steps are: • Convert to metres per hour by multiplying by 1600 (1 mile ≈ 1600 m) • Convert to metres per second by dividing by 3600 So overall, speed in ms-1 = speed in mph × 1600÷3600 (or × 4/9, if you cancel down) - enter this as a formula 3. Set up the third column as thinking distance - use a reaction of two thirds of a second 4. Set up the fourth column as braking distance - use the formula v2/13.17 5. Set up the fifth column as stopping distance - add the third and fourth columns You could also use your data to draw graphs - choose the xy (scatter) option to plot a "normal" graph

Car safety features

Analysing accidents

You need to be able to apply your knowledge of physics to understand how car safety features work.

You need to be able to use Newton's second law (F = ma) , the equations of motion and the idea that work done = force × distance moved to analyse accidents. This section concentrates on worked examples of this type.

Passenger cells & crumple zones

A car that was involved in an accident left skid marks stretching for 40m. From previous tests on the road, it is knwn that a skidding car decelerates at approximately 6 ms-2. (a) Estimate the speed of the car before it started to skid. (b) State any assumptions you have made in your calculation

crumple zone passenger cell

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The passenger cell is a rigid box which is likely to survive an impact The front of the car is the crumple zone. This is designed to squash up readily in a crash - so the distance in which the car stops is longer. Squashing the crumple zone uses up some of the kinetic energy of the car, which makes the crash less dangerous to the passenger. Air bags In a crash, the passengers are still moving forward after the car has stopped - they may be brought to a halt by hitting the front of the car. This abrupt collision with the steering wheel or windscreen is likely to cause injury. The purpose of an air bag is to bring passengers to a halt more slowly and gently. A sensor in the car tells the bag when to inflate - it inflates when there's a collision force on the car of a sufficient magnitude. The inflation is produced by a chemical reaction that releases a gas. A short time later, the gas dissipates through tiny holes in the bag - so the bag deflates. This is important to avoid passengers bouncing back off the bag.

Seat belts Like airbags, seat belts are intended to stop the passenger hitting the windscreen, steering wheel or any other hard part of the car, in a crash. They spread the stopping force over a wider area, and ensure it is exerted on more resilient parts of the body. They are also slightly stretchy, which also contributes to bringing passengers to a halt more slowly. The ends of the seatbelts are wound around an inertia reel. This allows you to pull the belt out slowly to fasten it, and causes it to retract if there is any slack in the system. However, if you try to pull the belt out sharply, the inertia reel clamps it. In a crash, the forward movement of the passengers will tend to jerk on the seatbelt - so the reel prevents the belt extending, thus holding the passengers in place.

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(a) Using 2as = v2 - u2 (because we know a and s and need u): 2(-6)(40) = 0 - u2 u = 21.9 ms-1 (b) We have assumed the car decelerates uniformly (as the equations of motion have been used) This sort of analysis allows us to determine whether a car was breaking the speed limit before an accident. A car does work of 4 × 105 J against a stopping force before coming to rest (a) Under normal conditions, the stopping force is 1000N. Calculate the distance travelled after the stopping force is applied (b) In an accident, the car is brought to rest in 10m. Calculate the stopping force exerted on the car in this case (c) Use your answers to help to explain why crumple zones help to reduce injuries in car accidents (a) Work done = force× distance moved 4 × 105 = 1000d d = 400m (b) 4 × 105 = 10F F = 4 × 104 N (c) If the car stops in a small distance, the forces exerted are large. The crumple zone increases the stopping distance, reducing the forces A car was travelling at a speed of 20ms-1. In an accident, it stops in a distance of 8m. (a) Calculate the deceleration of the car, assuming it is constant. (b) The car has a mass of 1000kg. Calculate the force exerted on the car to stop it. (a) 2as = v2 - u2 16a = -400 a = -25ms-2 (b) F = ma = -25000N

71 - Forces on Vehicles

Physics Factsheet

Typical Exam Question The "thinking distance" for a car travelling at speed v ms-1 is 0.667v m The "braking distance" for this car is 0.0759v2 m (a) Calculate the total stopping distance for the car when it is travelling (ii) 15 ms-1 [3] at (i) 10 ms-1 (b) A driver is driving on a winding road where the maximum visibility is 20 metres ahead. Explain why he should not drive at 15 ms-1[2] (a) Stopping distance =0.667v + 0.0759v2! (i) distance = 14.3m ! (ii) distance = 27.1 m! Braking distance is given by s = u2/2a !

Questions 1. Explain how airbags reduce injuries in car accidents

2. Explain why (a) Seatbelts are wide rather than narrow (b) Seatbelts are slightly elastic (c) The inertia reel is important in seatbelts,

3. By drawing a graph of stopping distance against speed, or otherwise, find the speed for which the stopping distance is (i) 30 metres (ii) 50 metres Use the data for reaction time and deceleration given on page 1

(b) His stopping distance is greater than the distance he can see ahead! So he could not stop in time if a hazard occurred.!

Typical Exam Question (a) Explain the meaning of "thinking distance" and "braking distance" [2]

4. Explain how (i) the weather (ii) the condition of the tyres affects the stopping distance of a car.

(b) Explain, using relevant equations, why braking distance is proportional to the square of the speed, but thinking distance only increases linearly with the speed of the car. [3] (a) Thinking distance is the distance travelled between the driver seeing a hazard (a reason to stop) and applying the brakes ! Braking distance is the distance travelled between the brakes first being applied and the vehicle coming to rest.!

5. When a car comes to rest, it must do work of 6 × 105 J. The normal stopping force applied is 1500N Calculate the distance travelled after the stopping force is applied

(b) The car travels at constant speed during thinking time ! So thinking distance is given by s = ut (t = thinking time)! Braking distance is given by s = u2/2a !

6. A car skids for 50 metres before an accident. Its speed before the accident was 60 miles per hour (a) Express the car's speed in metres per second (1 mile is 1.6 km)

Typical Exam Question The deceleration that can be exerted by a particular car under normal conditions is 5 ms-2 The driver of the car has a reaction time of 0.5 seconds - that is, he will start braking 0.5 seconds after seeing a hazard on the road ahead.

(b) Find the deceleration of the car (c) The car had a mass of 1200kg. Calculate the work required to stop the car.

The car is travelling at speed v ms-1. Find an expression in terms of v, for the stopping distance of the car. [3]

Acknowledgements: This Physics Factsheet was researched and written by Cath Brown. Curriculum Press,Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136 (c) Force = ma = 1200× 7.11 = 8533 N Work = force ×distance moved = 8533 × 50 (b) 2×a×50 = -26.672

Thinking distance = 0.5v ! 2× -5 × s = -v2 ! s = v2/10 ! So total stopping distance = 0.1v2 + 0.5v !

= 427000J (3 SF)

a = - 7.11 ms-2

6. (a) 60 mph is 60× 1.6 = 96 kmh-1 This is 96000 mh-1 This is 96000/3600 = 26.67 ms-1 5. 6 × 105 / 1500 = 400m 4. rain, ice and badly maintained tyres reduce the braking distance, but not the thinking distance 3. (i) 16.0 ms-1

(ii)

21.6 ms-1

2. (a) This reduces the pressure on the bodies of passengers (pressure = force/area) (b) This brings passengers to a halt more slowly (c) This allows the belt to be pulled out slowly, but stops it moving if it is tugged sharply. This stops the seatbelt extending in an accident, which keeps the passenger in place. Answers 1. They bring passengers to a halt more slowly, and also stop them "bouncing off" by deflating again.

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