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8.1.9 Estimating the effects of large accelerations

Large accelerations on the road

Definition

Acceleration

The rate of change of velocity.

Definition

Estimate

An approximate value of a quantity worked out from sensible, rounded values rather than from exact measurements.

  1. A large acceleration happens when the velocity of a vehicle or a person changes by a lot in a very short time.
  2. On the road the largest accelerations happen in collisions, when a vehicle stops in a small fraction of a second.
  3. Very hard emergency stops and very fast starts also produce large accelerations, though much smaller ones than a crash.
  4. To estimate the effects you work out three things in turn, the speed before the event, the acceleration during it and the force that produces it.
  5. Comparing an acceleration with the free-fall acceleration of 10 m/s210\ \text{m/s}^210 m/s2 shows how large it is, so 50 m/s250\ \text{m/s}^250 m/s2 is five times the acceleration of free fall.

Estimating the speed

Definition

Speed

The distance travelled by an object per unit time.

  1. The speed before an event can be estimated from the type of road, so about 13 m/s13\ \text{m/s}13 m/s in a 303030 mph zone and about 30 m/s30\ \text{m/s}30 m/s on a motorway.
  2. A speed given in mph or km/h\text{km/h}km/h must be converted into m/s\text{m/s}m/s before it is used in any equation.
  3. When a vehicle is filmed covering a known distance in a known time, its speed is the distance divided by the time.
  4. Accident investigators can also estimate a speed from the length of the skid marks, because a longer braking distance on the same surface means a higher speed.

Estimating the acceleration

  1. If the stopping time is known or can be estimated, the acceleration comes from a=v−uta = \dfrac{v-u}{t}a=tv−u​.
  2. If the stopping distance is known instead, the acceleration comes from v2−u2=2asv^2 - u^2 = 2asv2−u2=2as, rearranged to a=v2−u22sa = \dfrac{v^2 - u^2}{2s}a=2sv2−u2​.
  3. A car with a crumple zone that hits a solid wall typically stops in about 0.1 s0.1\ \text{s}0.1 s, while its front crushes by about half a metre.
  4. A person who hits a hard surface, such as a windscreen or the road, stops in only a few hundredths of a second.
  5. Collision decelerations are therefore tens or even hundreds of metres per second squared, many times the acceleration of free fall.
Example

A car hitting a wall

  • A car of mass 1200 kg1200\ \text{kg}1200 kg travelling at 13 m/s13\ \text{m/s}13 m/s hits a wall, and its crumple zone brings it to rest in about 0.10 s0.10\ \text{s}0.10 s.
a=v−ut=0 m/s−13 m/s0.10 s=−130 m/s2 a = \frac{v-u}{t} = \frac{0\ \text{m/s} - 13\ \text{m/s}}{0.10\ \text{s}} = -130\ \text{m/s}^2 a=tv−u​=0.10 s0 m/s−13 m/s​=−130 m/s2 F=ma=1200 kg×130 m/s2=156 000 N F = ma = 1200\ \text{kg} \times 130\ \text{m/s}^2 = 156\,000\ \text{N} F=ma=1200 kg×130 m/s2=156000 N
  • The car decelerates at about 130 m/s2130\ \text{m/s}^2130 m/s2, which is 131313 times the acceleration of free fall, and the force on it is about 160 000 N160\,000\ \text{N}160000 N.
Example

Stopping over a known distance

  • A car of mass 1200 kg1200\ \text{kg}1200 kg travelling at 20 m/s20\ \text{m/s}20 m/s runs into a crash barrier, which bends and stops it over a distance of 2.0 m2.0\ \text{m}2.0 m.
a=v2−u22s=02−2022×2.0 m=−100 m/s2 a = \frac{v^2 - u^2}{2s} = \frac{0^2 - 20^2}{2 \times 2.0\ \text{m}} = -100\ \text{m/s}^2 a=2sv2−u2​=2×2.0 m02−202​=−100 m/s2 F=ma=1200 kg×100 m/s2=120 000 N F = ma = 1200\ \text{kg} \times 100\ \text{m/s}^2 = 120\,000\ \text{N} F=ma=1200 kg×100 m/s2=120000 N
  • A concrete wall might stop the same car in only 0.5 m0.5\ \text{m}0.5 m, giving a deceleration and a force four times larger.

Estimating the force on a person

Definition

Resultant force

The single force that has the same effect on an object as all of the forces acting on it combined.

Definition

Weight

The force acting on an object because of gravity, measured in newtons.

  1. Once the acceleration is known, the resultant force on any object comes from F=maF = maF=ma, using the mass of that object.
  2. The force on the whole car uses the mass of the car, while the force on a passenger uses only the passenger's mass of about 70 kg70\ \text{kg}70 kg.
  3. Comparing the force with the person's weight, which is about 700 N700\ \text{N}700 N, shows how severe it is.
  4. The larger the force compared with the weight, and the smaller the area of the body it acts on, the more serious the injuries are likely to be.
  5. Increasing the stopping time, with a crumple zone, seat belt or airbag, reduces the acceleration and so reduces the force on the body.
  6. Large accelerations when speeding up are much smaller than in a crash, but they still need a large force from the seat on the driver's back.
Example

Force on a passenger

  • A 70 kg70\ \text{kg}70 kg passenger in the car that hit the wall is held by a seat belt that stretches, so the passenger stops in about 0.20 s0.20\ \text{s}0.20 s.
a=v−ut=0 m/s−13 m/s0.20 s=−65 m/s2 a = \frac{v-u}{t} = \frac{0\ \text{m/s} - 13\ \text{m/s}}{0.20\ \text{s}} = -65\ \text{m/s}^2 a=tv−u​=0.20 s0 m/s−13 m/s​=−65 m/s2 F=ma=70 kg×65 m/s2=4550 N F = ma = 70\ \text{kg} \times 65\ \text{m/s}^2 = 4550\ \text{N} F=ma=70 kg×65 m/s2=4550 N
  • The force of about 4600 N4600\ \text{N}4600 N is roughly 6.56.56.5 times the passenger's weight of 700 N700\ \text{N}700 N.
  • Without the seat belt, a passenger hitting the dashboard in 0.02 s0.02\ \text{s}0.02 s would experience a force ten times larger, about 46 000 N46\,000\ \text{N}46000 N.
Example

A very fast start

  • A sports car reaches 27 m/s27\ \text{m/s}27 m/s from rest in 3.0 s3.0\ \text{s}3.0 s, and its driver has a mass of 70 kg70\ \text{kg}70 kg.
a=v−ut=27 m/s−0 m/s3.0 s=9.0 m/s2 a = \frac{v-u}{t} = \frac{27\ \text{m/s} - 0\ \text{m/s}}{3.0\ \text{s}} = 9.0\ \text{m/s}^2 a=tv−u​=3.0 s27 m/s−0 m/s​=9.0 m/s2 F=ma=70 kg×9.0 m/s2=630 N F = ma = 70\ \text{kg} \times 9.0\ \text{m/s}^2 = 630\ \text{N} F=ma=70 kg×9.0 m/s2=630 N
  • The seat pushes the driver forwards with a force close to their own weight, which is why they feel pressed back into the seat.
Common Mistake
  • Do not use the mass of the car when the question asks for the force on a passenger, because the force on each object depends on its own mass.
  • Do not forget to convert a speed in mph or km/h\text{km/h}km/h into m/s\text{m/s}m/s before using it in an equation.
  • Do not give an estimate to many significant figures, because the stopping time and the mass are only approximate.
Exam technique

Estimation questions

  • Set out the estimate in stages, with the speed first, then the acceleration, then the force, and show each value you have assumed.
  • Use a=v−uta = \dfrac{v-u}{t}a=tv−u​ when a stopping time is given and v2−u2=2asv^2 - u^2 = 2asv2−u2=2as when a stopping distance is given.
  • Round the final answer sensibly, give its unit and, if asked how dangerous the acceleration is, compare the force with the person's weight.
Self review
  • Estimate the speed in m/s\text{m/s}m/s of a car travelling at 303030 mph.
  • Calculate the deceleration of a car that stops from 20 m/s20\ \text{m/s}20 m/s in 0.10 s0.10\ \text{s}0.10 s.
  • Estimate the force on a 70 kg70\ \text{kg}70 kg passenger who decelerates at 100 m/s2100\ \text{m/s}^2100 m/s2.
  • Explain why the force on a passenger is smaller if they stop over a longer time.
  • State the equation you would use to find a deceleration when only the stopping distance is known.
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8.1.9 Estimating the effects of large accelerations Revision Guide

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Revision notes for OCR GCSE Physics 8.1.9 Estimating the effects of large accelerations: explanations and worked examples.

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