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GCSE Physics notes: Reaction times, stopping distances, kinetic energy calculations

5. Reaction times and stopping distances e.g. road vehicles and problem solving using Newton's 2nd law equation and kinetic energy calculations

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

What is the formula for stopping distance?  What factors affect thinking distance?

What factors affect braking distance?   What is the relationship between braking distance and kinetic energy?   Can you devise a simple experiment to measure somebody's reaction time?

(a) Introduction - stopping distances and the speed of road vehicles

(b) How to calculate thinking distances and braking distances from velocity - time graphs

(c)

(d)

(e) Graphical analysis of stopping distances, speed and kinetic energy of a moving car

(f) More on the physics of vehicle braking and kinetic energy

(g)

(h)

(i) Simple reaction time experiments

(a) Introduction - stopping distances and the speed of road vehicles

When driving a car you obviously need to be alert to any sudden change in your situation, particularly if you have to do an emergency brake to come to a halt.

In this situation you want to stop the car (or any other road vehicle) in the shortest time as possible to effect an appropriate emergency stop!

This means applying the maximum force on the brake pedal.

The longer it takes for you to react and longer it takes to come to a halt, the greater the risk of crashing into an object in your path. Everybody's 'thinking' reaction time to a situation requiring a quick physical response is different, though typically in the range 0.2 to 0.8 seconds. In biology you may have studied the nervous system including the reflex arc.

The distance it takes to stop a road vehicle in an emergency situation is given by the following formula:

STOPPING DISTANCE = THINKING DISTANCE + BRAKING DISTANCE

The thinking distance is how far you travel during your reaction time, which is the time interval from you perceiving a hazard and actually starting to take action e.g. apply the brakes.

The braking distance is the actual distance you travel from when you first apply the brakes until you come to a halt.

The stopping distance is the total time required from initial visual stimulus to actually stopping moving.

The chart above gives typical or average values for thinking distances, braking distances and stopping distance and are quoted from the UK Highway Code guidance booklet.

You can see the thinking distance is quite an appreciable portion of the total stopping distance, particularly at lower speeds, BUT, look how dramatically the total stopping distance increases with increase in speed.

These values should be doubled for wet roads and multiplied by 10 for ice coated roads. Snow would be somewhere in between, but where?, so just take great care in driving in any of these adverse driving conditions.

Later in this page I've used this data to do some graphs and calculations relating stopping distance to speed and the kinetic energy of a car.

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(b) How to calculate thinking distances and braking distances from velocity - time graphs

Graphs 1a

You have probably encountered velocity time graphs by now, so you should know that the area under a section of a velocity- time graph is equal to the distance travelled in that section (in terms of units m/s x s = m).

The graphs assume the same car and driver so that the deceleration on maximum braking is the same, which is why the negative gradient is the same value on both graphs.

The graph on the left of 1a shows an initial situation of a driver's quicker response time travelling at a lower speed.

Rectangular area A1 = initial velocity v1 x reaction time t1 = thinking distance

The area A1 is equal to the thinking distance, that is the distance the vehicle travels in the time it takes the driver to respond to a situation and starts to apply the brakes.

Right angled triangular area A2 = ½  x initial velocity v1 x braking time t2 = braking distance

The area A2 is the braking distance, that is the distance the vehicle travels from its maximum initial speed, when braking starts, until it comes to a halt.

The total area = A1 + A2 = stopping distance

The graph on the right of 1a shows a driver's slower reaction and the vehicle is moving at a greater speed.

This means two factors have been changed to emphasise how stopping distance is so easily and dramatically  increased.

So v2 > v1 and times t1 and t2 are both increased, so both areas A1 and A2 are increased.

The purple shaded areas indicate the increase in thinking distance A1 and braking distance A2.

This might mean lack of care and attention e.g. tired and not concentrating on the speed limit.

Rectangular area A1 = initial velocity v2 x reaction time t1 = thinking distance

Right angled triangular area A2 = ½  x initial velocity v2 x braking time t2 = braking distance

So, both area A1 and A2 are greatly increased, increasing the likelihood of an accident if driving carelessly!

The total area = A1 + A2 = stopping distance, and much greater than before.

If you have followed the above logical arguments, you should be able to interpret the graphs if only one factors was changed.

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(c) Factors affecting the thinking distance (ultimately affecting stopping distance too)

Speed is the first obvious factor.

The faster you are going, the further you will travel for the same 'best' reaction time you can manage, so the greater the thinking distance, which you can do nothing about.

You can only keep this to a minimum by being fully alert and able to respond as fast as your body is able to.

The effects of tiredness and alcohol will affect your alertness and increase your response time and thinking distance.

There other factors too.

Are you talking to somebody else in the car, are the kids being silly?

Even using a mobile phone legally with a hand-set, is still potentially a distraction.

Poor visibility e.g. fog or smoke, will delay you spotting a hazard and reacting to it, so effectively increasing your thinking time.

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(d) Factors affecting the braking distance (ultimately affecting stopping distance too)

Again, speed is the first obvious factor.

The faster you are going the more kinetic energy has to removed from the car's kinetic energy store. At a constant rate of braking force, it will take longer the greater the speed, because more kinetic energy has to be converted to heat energy in the brake pad and disc system.

This is shown on the right (brake pads P in contact with disc D).

All of the factors discussed here become particularly crucial in an emergency braking situation or you suddenly find yourself too close to the car in front.

The greater your speed, the greater the stopping distance and the greater distance you should allow between one vehicle and another e.g. the two chevron distance for 70 mph you see on some sections of a motorway.

However good your brakes are, its no good being too close to another vehicle i.e. well within the stopping distance, if you are to avoid an accident if the vehicle in front does an emergency brake or the traffic head rapidly comes to halt!

Speed limits aren't simply about speed reduction, they are also about reducing both the stopping distance where higher speeds are considered hazardous for a particular section of road. This for the safety of road users and pedestrians e.g. 20 mph in narrow streets in built up areas where there are likely to be many people walking and crossing roads.

The condition of the tyres: Tyres are designed to give the maximum road grip and expel water from under the tyres on wet roads. If the tyres are worn (bald or little tread left), there is less grip and the vital friction and water expelling function to slow the vehicle down are reduced and so increasing the braking distance and the chance of skidding. Also, the tyres should contain enough air to give the correct operating pressure.

The effectiveness of the brakes: If the brakes are not well maintained, braking function may be impaired. Brake pads might be worn or a leak in the hydraulic brake system can be a source braking impairment. Are the brakes balanced so that you slow down in a straight line - this is point applies to the condition of the tyres too.

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(e) Graphical analysis of stopping distances, speed and kinetic energy of a moving car

see calculations

Diagram KEY: KE = kinetic energy (J), m = mass (kg),   u = initial velocity (m/s),   v = final velocity (m/s),   s = speed (m/s)

a = acceleration or deceleration (m/s2),   W = work done (J),   F = force (N),   d = distance (m)

Graph 1b

The graph 1b above takes the thinking distance, braking distance and stopping distance data and plots them against the typical speed of a road vehicle.

Obviously, all the distances increase with increase in speed, but note two other very important points.

You should notice ...

(i) two of the graphs curve upwards, so there is a sort of 'accelerating' effect of speed on the braking distance and overall stopping distance (the latter is due to the increase in braking distance).

Stopping distance and braking distance are not proportional to speed, and crucially,  the braking distance is proportional to speed squared. This means the stopping/braking distances increase faster than the increase in speed.

e.g. doubling speed quadruples the braking distance (2 ==> 22 = 4) and trebling speed increases the braking distance nine times (3 ==> 32 = 9).

The thinking distance is roughly proportional to speed, the graph is ~linear and does not curve upwards. This is because your response time, if fully alert, is pretty constant, so if your speed doubles, you just go twice as far in the same response time.

(ii) and if you examine the graph or data carefully, you can see that doubling the speed quadruples the braking distance.

This means by doubling your speed, approximately quadruples the stopping distance, obviously something you need to bear in mind the faster you drive.

Doubling speed quadruples braking distance and trebling speed increases it nine times! (see the REMINDER below)

This is discussed further and is related to the formula for kinetic energy KE = ½mv2.

By doubling the speed, you quadruple the kinetic energy of the car, hence you have quadrupled the kinetic energy to be removed by braking (because KE v2). See graphs 2 and 3 and notes below.

Therefore, on doubling the speed, for a constant braking force, you have four times as much KE to remove and will need four times the distance to remove it.

For more on kinetic energy calculations see Kinetic energy store calculations

• REMINDER: The kinetic energy of a moving object can be calculated using the equation:

• kinetic energy (KE) = 0.5 × mass × (speed)2

• Eke = 1/2 m v2

• kinetic energy, Eke, in joules, J; mass, m, in kilograms, kg; speed/velocity, v, in metres per second, m/s

• Note that by doubling the speed you quadruple the kinetic energy! (e.g. 22 : 42 is 4 : 16 or 1 : 4)

Question to illustrate some of the ideas above and using the chart below.

When travelling at 20 mph a driver's thinking distance is 6.0 m and the braking distance is 6.0 m.

(a) What is the stopping distance?

stopping distance = thinking distance + braking distance = 6.0 + 6.0 = 12.0 m

(b) Estimate the total stopping distance at 40 mph (scale facto 2).

If the thinking distance is 6 m at 20 mph, it will be double that at 40 mph, 6 x 40 / 20 = 12 m.

From the KE argument and KE v2, the braking distance increases by the square of the scale factor.

So the braking distance 6 x 22 = 24 m

Therefore the stopping distance is 12 + 24 = 36 m (check on the chart)

(c) Estimate the total stopping distance at 80 mph (scale factor 4).

If the thinking distance is 6 m at 20 mph, it will be quadruple that at 40 mph, 6 x 80 / 20 = 24 m

The braking distance increases by the square of the scale factor.

So the braking distance 6 x 24 = 96 m

Therefore the stopping distance is 24 + 96 = 120 m (not on the chart)

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(f) More on the physics of vehicle braking and kinetic energy

The mechanical process of braking primarily relies on friction between the brake pad and a steel disc (shown on the right). When you press the brake pedal the hydraulic system pushes pads onto the surface of the disc causing work to be done due to the resistive forces between the surfaces.

The resulting friction effect transfers energy from the car's kinetic energy store to the thermal energy store of the braking system which is eventually dissipated to the environment's energy store.

The friction does cause the brakes to heat up - the brake pads and disc must be able to withstand a high temperature - both are made of high melting alloys.

A little of the KE is lost as sound.

If the wheel tyres skid on the road, friction will generate thermal energy and the road and tyre increase in temperature.

Eventually all the kinetic energy of the road vehicle is dissipated to the thermal energy store of the surroundings.

So when the work is done between the brakes and the wheel discs kinetic energy is converted to thermal/heat energy.

The faster a vehicle is going, the greater its kinetic energy store and more work must be done to bring the car to a halt.

It also means a greater force must be applied to bring the vehicle to a halt within a certain braking/stopping distance.

The greater the braking force, the greater the deceleration.

Big decelerations can be dangerous because the brakes may overheat affecting their action AND there is a much greater chance of skidding, particularly if the road surface is slippery due to conditions already described.

To put the point about kinetic energy into context, study graph 2 below.

Graph 2

Graph 2 shows how the kinetic energy of a road vehicle (e.g. a car of 1200 kg) varies with its speed.

You can see that by doubling the speed, you quadruple the kinetic energy of the car, hence you have quadrupled the kinetic energy to be removed by braking.

This is because KE = ½mv2. Its the speed2 term that gives this crucial mathematical importance.

Assuming uniform deceleration and uniform decrease in the rate of reducing kinetic energy, means the braking distance is a function of kinetic energy and speed2. See graph 3 now.

Graph 3 shows the linear relationship between the kinetic energy of the car and braking distance (using the UK Highway Code data and a 1200 kg car).

Graph 3

This is a result of KE = ½mv2 and the braking distance data assumes uniform deceleration and uniform decrease in the rate of reducing kinetic energy due to the friction of the brakes.

As already mentioned, the braking distance increases faster than the speed.

The total work done to stop a road vehicle is equal to the initial maximum kinetic energy of the vehicle.

Work done to halt vehicle = total KE of vehicle = braking force x braking distance

W = F x d = KE = ½mv2   (in a nutshell !)

W = work in J to come to a halt, and all of the work is done by the brakes (assuming no skidding) via friction from the vehicles KE store to the thermal energy store of the brakes and environment

F = braking force in N (assumed to be constant for the vehicle brakes),

d = braking distance in m, m = mass of vehicle in kg, v = speed of vehicle in m/s

If you skid on a dry road, the rubber left on the road tells you the tyres were doing a bit of braking work too!

If we assume a constant braking force (maximum push on brake pedal) and since the kinetic energy of the car is proportional to speed2, then the braking distance is proportional to the initial kinetic energy of the car.

That's what the work done equation says for a constant braking force:

KE BD and so does the graph.

An extra consequence: If your car is full of people or a lorry is fully loaded, then the kinetic energy at a given speed is greater than if the vehicle only contained the driver. Therefore, with extra mass in the vehicle, extra distance should be allowed for your braking distance because of the extra kinetic energy.

Examples of typical masses for road vehicles:

cars 1000 - 1500 kg;  large van/single decker bus ~9 000 -10 000 kg; loaded lorry ~30 000 - 40 000 kg.

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(g) Health and safety issues related to collisions involving road vehicles and cyclists

(motor bikes, cars, lorries, buses etc.)

Introduction

Large decelerations (rapid slowing down) of objects (cars crashing or people falling and hitting the ground) involve substantial forces and can obviously cause injury.

Why? Large decelerations require a large resistive force. Recall Newton's 2nd law of motion equation ...

F = ma, to create a large acceleration a, you need a relatively big force F, irrespective of the mass m,

also, the bigger the mass m, the larger the force F needed for a given deceleration.

In principle, the force experienced by an object can be reduced by reducing the deceleration ('slower' slowing down).

Recall: acceleration = change in speed / time taken, a = ∆v/∆t, increase ∆t to decrease a

In terms of momentum, you are trying to change the momentum over as longer time as possible, to minimise the force involved.

In the next section we apply these ideas to design safety features that increase collision times - the time from the initial impact of an object with an obstacle to the object coming to a halt (∆t in terms of above equations) i.e. to reduce the rate of deceleration.

You need to know about things like air bags and safety belts in cars, crumple zones at the front and rear of cars, safety helmets for cycling.

Applying the physics of forces to safety design

In a collision between a road vehicle and a stationary object the normal contact forces between them will cause work to be done.

The collision will cause energy to be transferred from the vehicles kinetic energy store to several other energy stores.

The thermal energy (impact friction) and elastic potential energy ('crunching' effect) stores of the two objects will increase and some of the kinetic energy will end up as sound.

When everything has 'settled down' after the crash, theoretically, all the kinetic energy store of the moving vehicle eventually ends up increasing the thermal energy store of the environment.

You can build safety features into the design of road vehicles, and, where appropriate, safety clothing.

What you are trying to in most cases is to slow down the deceleration - increasing the collision time or absorbing the kinetic energy of any rapid deceleration and in doing so minimise the force a person's body experiences. A rapid impact produces an extreme deceleration - much more so than even emergency braking.

Its all about minimising injury to people in a rapid change of motion situation.

In terms of physics, its all about absorbing impact energy and increasing the deceleration time - minimising the a in F = ma!

From Newton's 2nd law of motion: F = ma, so for a given mass m, if you can make a the deceleration smaller, the decelerating force F is also reduced and minimises body impact and injury.

Wearing a seat belt reduces the force of impact of the deceleration.

On collision or in emergency braking, the seat belt stretches a little, increasing your deceleration time and decreasing the force your body experience against the seat belt. The rate of change of momentum is reduced (F = ∆mv/∆t)

Fast acting air bags, cushion your body from a violent impact, they also increase deceleration times and reduce the force your body experiences. Again, the rate of change of momentum is reduced (F = ∆mv/∆t)

The air bags rapidly expand and are then compressed as a car occupant crashes into it.

The compression takes a longer time than if you crashed into the dashboard of a crushed car, or even if you were compressed excessively confined in your seat belt.

A car body can have crumple zones, built into the design of the car body, in both the front and back to absorb the kinetic energy of any high impact. This increases the deceleration time, thereby decreasing the force your body experiences.

The photographs of a (faked) moderately violent crash of the car into a brick wall gives you an idea of what a 'crumple zone' is all about.

You would see similar damage to the rear of your car (2nd crumple zone) if someone runs into the back of you.

Cycle helmets and crash helmets

Helmets worn by cyclists or motor bike riders (motorcyclists) have an inner lining of foam (or other energy absorbing material) to cushion the head on impact.

The foam increases the time before your head stops moving due to the impact.

The smaller deceleration over a greater period of time, decreases the force of impact your head experiences.

 BIKE HELMETS Everything is designed with safety (and comfort) in mind. The main safety features of the motor cycle crash helmet are the hard protective outer shell and the 'soft' impact energy absorbing liner. The comfort fitting padding 'foam' will absorb kinetic energy on impact. Image from the CALIFORNIA MOTORCYCLIST SAFETY PROGRAM and supported by the California Highway Patrol The scheme advises motorcycle riders with helmets that do not match all the safety design features illustrated, should change helmet!

When out walking, came across a motorcycle couple who kindly allowed me to take photographs. Both survived a serious accident, but once the crash helmet has been in an impact situation, it must be replaced. You can clearly see all the features described in the diagram above.

So, teenage motorcyclists, buy the safest helmet, it might cost more, but without the best helmet, it might cost you even more.

Research is always going on to develop new materials to increase the performance of safety features, whether it be car bodies or helmets.

The same ideas apply to safety in play areas for children and safety in sports like gymnastics

Playground equipment is sited on safety mats that absorb the force of impact when a child falls on them.

They be made of rubber or foam materials.

The idea of this 'cushioned' play area flooring is to increase the impact time by using material that compresses under impact, which cannot happen with a hard surface.

If gymnasts need to make a landing from a piece of apparatus they need to land on a cushioned surface to lessen the impact force the legs experience and avoid injury.

The safety mats are particularly needed when learning new routines where errors and accidents are more likely to happen.

In both competition and practice, the use of mats is now mandatory on most events and gymnasts may use an additional landing mat, without deduction as long as they land within a specified distance.

Even footballers wear humble shin pads to protect their legs from hard tackles!

The thick layer of material absorbs the energy of impact from the 'tacklers' leg or boot, increasing the impact time and reducing the impact force.

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(h) Some advanced calculations on braking force and kinetic energy

Diagram KEY: KE = kinetic energy (J), m = mass (kg),   u = initial velocity (m/s),   v = final velocity (m/s),   s = speed (m/s)

a = acceleration or deceleration (m/s2),   W = work done (J),   F = force (N),   d = distance (m)

Q1 Suppose a car of 1200 kg is travelling at 18 m/s (~40 mph) and has to do an emergency stop with a hazard 30 m ahead.

(a) Calculate the deceleration of the car and (b) the braking force involved.

(a) First use the motion equation v2 - u2 = 2ad to calculate the deceleration.

where v = final velocity, u = initial velocity, a = acceleration (∆v/∆t), d = distance travelled

Assuming uniform deceleration and v = 0 (comes to halt), u = 18 m/s, d = 30 m

v2 - u2 = 2ad, 0 - 182 = 2 x a x 30

60a = -324, therefore a = -324/60 = -5.4 m/s2 (note the negative sign for deceleration)

(This is easier to do if your given the braking time, so can just use a = ∆v / ∆t, which I did in the previous section comparing a car and a heavy goods vehicle, and called it Q2)

(b) You then use Newton's 2nd Law equation F = ma,

where F = decelerating braking force, m = mass of car,

a = deceleration of car = change in speed / time taken

Substituting into the equation (and you can ignore the -ve acceleration sign here, but NOT above)

F = ma = 1200 x 5.4 = 6480 N

Comment: That's why your body is thrown forward. The deceleration is just over half the value of the acceleration you experience due to the Earth's gravitational field. If you are involved in a high speed impact the force can be much greater and hence destructive on you and the car!

See section on

Q2  A small domestic 1000 kg car (1 tonne) with two axles at 60 mph (26.84 m/s)

will have a kinetic energy = 0.5 mv2 = ½ x 1000 x 26.842 3.6 x 105 J (360 kJ, 3 s.f.)

A heavy articulated goods vehicle of 6 axles may weigh, with a full load, up to 43 000 kg (43 tonnes) at 60 mph (26.84 m/s)

will have a kinetic energy = 0.5 mv2 = ½ x 43,000 x 26.8421.55 x 108 J (15 500 kJ, 3 s.f.)

Now, both of these vehicles have to be able to stop in the same safe distance in an emergency.

The two axle car will have four sets of brake pads.

The six axle goods vehicle will have twelve sets of brake pads, three times as many as the car.

This means to stop in the same safety distance, the braking force exerted by each set of pads in the goods vehicle must be much greater than for the car.

At 50 mph (22.37 m/s) suppose the safe braking distance is 38 m.

We can then calculate the total braking force needed to stop in three seconds.

(i) for both vehicles deceleration a = ∆v / ∆t = 22.37 / 3 = 7.457 m/s2

(ii) F = ma from Newton's 2nd Law, force in newtons, mass in kg, deceleration in metres per second2

For the car: F = 1000 x 7.457 = 7 460 N (3 s.f.),

that is 1865 N braking force per set of four brake pads.

For goods vehicle: F = 43 000 x 7.457 = 321 000 N (3 s.f.).

that is 26750 N braking force per set of brake pads.

This means the heavy goods vehicle brake pads must generate over 14 x the braking force of the car.

(For those expert in road vehicle physics, I do appreciate these are simplified calculations)

For more on F = ma calculations see Newton's Second Law of Motion and momentum calculations

Q3 Suppose a car travelling at 30 m/s (~70 mph) has to make an emergency stop to avoid a hazard.

If the mass of the car is 1500 kg, the braking force of the car is 6000 N and the tired driver's reaction time is 1.5 seconds, calculate the following:

(a) Calculate the thinking distance of the driver (s = speed (m/s), d - distance (m), t = time (s))

s = d / t, d = s x t = 30 x 1.5 = 45 m = thinking distance

(b) Calculate the initial kinetic energy of the car (m = mass of car in kg, v = speed of car (m/s)

KE = ½mv2 = 0.5 x 1500 x 302 = 675000 = 6.75 x 105 J = initial KE of car

(c) Calculate the braking distance to halt the car  (W = braking work done (J), d = braking distance (m)

Work done in braking the car must equal the kinetic energy of the car (see Graph 3 discussion)

W = F x d = KE = ½mv2 = 6.75 x 105 J

W = F x d,  d = W / F = 6.75 x 105 / 6000 = 113 m = braking distance (3 s.f.)

(d) Calculate the stopping distance of the car

stopping distance = thinking distance + braking distance

= 45 + 113 = 158 m = stopping distance

Q4 See reaction time experiment

Q5 A 1500 kg 4x4 car travelling at 18.0 m/s (~40 mph) veers off the road, without slowing until hitting and demolishing a brick wall.

If it took 0.200 seconds to demolish the wall, calculate the following ...

(a) What is the initial kinetic energy of the car?

KE = ½mv2 = 0.5 x 1500 x 182 = 243 000 = 2.43 x 105 J

(b) What work is done on the wall and car in bringing the car to a halt?

2.43 x 105 J because all the kinetic energy of the car has to be removed.

(c) What happens to the kinetic energy of the car after impact?

The kinetic energy store of the car is reduced to zero and the energy is converted into heat (by compression or friction) and some sound energy (which will end up as heat too). So the thermal energy store of the wall, car and surrounding air is increased.

(d) Calculate the rate of deceleration

Deceleration = change in speed / time taken = ∆v / ∆t = (0 - 18) / 0.2 = -90 m/s2

(e) What is the decelerating force acting on the car?

From Newton's 2nd Law: F (N) = m (kg) x a (m/s2)

Decelerating force = 1500 x -90 = 135 000 = -1.35 x 105 N

The force (from the wall) is negative because it is acting in the opposite direction to the motion of the car.

If the car had braked in time, the decelerating force would be positive (in every sense of the word!).

Q6 Imagine a car of 1000 kg travelling at 20 m/s doing an emergency stop in a distance of 25 m - the braking distance.

Calculate the average braking force produced by the driver when pressing on the brake pedal.

To solve this question you to use several formulae.

(a) Calculate the kinetic energy of the car.

KE = 0.5 mv2 = 0.5 x 1000 x 202 = 200 000 J

(b) What work must be done to bring the car to a halt? Explain your answer.

If the kinetic energy of the car is 200 000 J, then 200 000 J of work must be done to bring the KE of the car to zero i.e zero velocity.

(c) Calculate the average braking force required.

Work (J) = force (N) x distance (m)

work = 200 000 J and the braking distance was 25 m

force = work / distance = 200 000 / 25 = 8000 N average braking force.

Q7  A van of mass 2000 kg veers off the road at 30 m/s and becomes stationary after hitting a stone wall.

(a) If the impact force on the van is 48 000 N, calculate the stopping time.

F = m∆v / ∆t, substituting

48 000 = 2000 x (30 - 0) / ∆t

48 000 = 60 000 / ∆t

∆t = 60 000 / 48 000 = 1.25 s

(b) Explain how a safety belt and an inflating air bag can save the drivers life.

On impact, the driver's body is accelerated forwards.

(i) The safety seat belt stretches sufficiently to reduce the rate of change of momentum - increasing the deceleration time.

(ii) The 'soft' inflated airbag also reduces the rate of change of momentum and absorbs kinetic energy when the driver's body hits it.

Q8 A 20 000 kg road vehicle comes to an emergency halt.

A uniform braking force of 8000 N is applied by the driver until the vehicle comes to a halt in a distance of 20 m.

(a) Calculate the speed of the vehicle just before the brakes were applied.

Work done in braking = braking force x distance brakes applied = 8 000 x 20 = 160 000 N

The total work done in braking = the kinetic energy of the vehicle at the instant the brakes are first applied.

KE = 0.5 mv2,  rearrangement gives v = √{(KE / (0.5 x m)}

v = √{(160 000 / (0.5 x 20 000)} = 4 m/s

(b) What are the major energy store transfers taking place?

The kinetic energy of the vehicle is mainly converted, via friction, to increase the thermal energy store of parts of the vehicle and the surrounding air or road.

For more on kinetic energy calculations see

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(i) Simple reaction time experiments

But, can be accompanied by some moderately complicated calculations!

Your reaction time to a situation may be typically 0.2 to 0.8 seconds when fully alert. However your reaction time can be affected by  tiredness, feeling unwell, drugs, alcohol, in other words anything that affects the speed of your brain function.

See

You can conduct quite simple experiments to test your reaction time to a particular situation. However, since the reaction time is too short, a stopwatch is no good, but there are ways of measuring your reaction time indirectly by making other measurements from which you can calculate your reaction time.

(a) Computer screen - where you respond as quickly as possible to something appearing on the screen.

In this situation, the computer software generates something up on the screen and automatically times your response by monitoring your contact with the keyboard or by clicking the mouse.

I've quickly written an extremely simple computer programme to test your response to a X appearing on the screen.

: It probably only works on Microsoft platforms, and maybe not all of them?

Your anti-virus protection might query it, because it is a .exe file, but its written with compiled BBC BASIC and should not pose any threat. Unfortunately I never learned to write in a multi-platform professional computer programming language, but I'm not exactly short of website projects!

(b) A simple physical response test - the falling ruler drop test

You get someone to hold a ruler vertically, with thumb and first finger, above someone else's hand, who is ready to catch it with their thumb and first finger.

First image on the right. The ruler should be held at the top of the scale and steady hands from both people.

The catching person should have the middle of their thumb and finger adjacent to zero on the cm scale - squat down to make sure you are reading the scale horizontally.

Then, without warning, the person holding the ruler, lets go of it. The second person has to react as fast as possible and catch the dropped ruler between their thumb and first finger.

Second image on the right. The longer the distance, the slower your reaction time!

When caught, you then read how far the ruler as fallen by taking the reading, to the nearest centimetre, from where the middle of their thumb and finger are.

You repeat the experiment a number of times to get an average, but its not a particularly accurate experiment.

You need to have steady hands and not let the ruler wobble about or fall at an angle other than vertical. You should also use the same ruler and the same people dropping the ruler and catching it (fair test criteria), though, obviously, you can compare one person's results with another.

The slower your response time, the further the ruler falls before being caught. You might repeat the experiment by having some background distractions - a group of people talking nearby, or somebody trying to engage you in conversation or music.

Q4 You can then do some 'nifty' calculations to actually obtain a real response time - so you using indirect data to get the response time.

It involves a two stage calculation.

Suppose the ruler is caught after an average fall of 25 cm.

(i) You use the equation v2 - u2 = 2ad, to calculate the final velocity  (more calculations using this equation)

v = final velocity (m/s), u = initial velocity (m/s), a = acceleration = 9.8 m/s2 (gravitational acceleration),

and d = distance fallen (m)

Since u = 0 and d = 25/100 = 0.25 m

v2 - 0 = 2 x 9.8 x 0.25 = 4.9

v = √4.9 = 2.214 m/s (its not this accurate, but we'll leave the s.f. until the end)

(ii) We can now use the acceleration formula to calculate the response time.

a = ∆v / ∆t, where a = acceleration (9.8 m/s2), ∆v = change in velocity (m/s) and ∆t = response time

Therefore : 9.8 = 2.214 / ∆t,    ∆t = 2.214 / 9.8 = 0.23 s (2 s.f.)

So, on average, the response time was about a quarter of a second.

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