P3 Physics Forces for transport

Speed  Miles per hour  Kilometres per hour  Metres per second  Centimetres per second  Kilometres per second  So what does the “per” mean?

“Per” means “divided by”  So kilometres per hour is the miles you did divided by the time it took.  There is a rule:  Speed = distance ÷ time  Or “average speed is distance over time”  Sometimes SIDOT  Has to be “average” because most things can’t keep to exactly the same speed all of the time.

Examples:  Travel 100 km in 2 hours, average speed is therefore:  100 ÷ 2 = 50  The unit for this is km/h  Travel 1000 km in 25 hours, average speed is therefore:  1000 ÷ 25 = 40  The unit for this is km/h

 Travel 1 m in 5 hours, average speed is therefore:  1 ÷ 5 = 0.2  The unit for this is m/h but you wouldn’t usually use that  Travel 200 m in 20 seconds, average speed is therefore:  200 ÷ 20 = 10  The unit for this is m/s

Try these:  What is the average speed of an object that travels:  10 metres in 5 seconds  100 metres in 5 seconds  200 metres in 5 seconds  10 metres in 5 hours  300 metres in 50 seconds  Don’t forget the units

The Triangles  If you cover the one you don’t know, the calculation is shown by the other two  This changes the subject of the equation Dist Speed x t

Acceleration  This is how much an object’s speed changes in a certain time  The units are always metres per second per second, written as m/s/s or m/s 2  As an equation: acceleration = final speed – start speed time taken to change  If the final speed is less than the start speed, then the object has decelerated – negative acceleration

Force and acceleration  It is our experience that a heavy object needs more force to get it moving than a light one. Stopping a heavy object, at the same speed, takes more force.  We also know that a lighter object will accelerate (and then move) faster than a heavy one.  This comes as one equation:  F = ma - force is mass times acceleration

What do we know so far?  Speed = distance  time  Distance – time graphs  Acceleration = (change in speed)  time  Force = mass x acceleration  Speed – time graphs  Work = force x distance  Power = work  time Dist Speed x t Work F x d Work P x t Change in speed A x t Force m x a

e.g.  200km in 4 hours is a speed of 50 km/h  m/s in 5 seconds is an acceleration of 20 m/s/s (m/s 2 )  Moving an object 5m with a force of 5N is 25 J of work (energy)  Doing 100 J of work in 4 seconds is 25 Watts.

Terminal velocity  When an object is accelerated by a force, it gets faster. (depends on the force, of course)  But as it goes faster, the friction and (often) air drag get bigger  So its speed reaches a limit, called terminal velocity.

Continued…  When an object has reached terminal velocity, the forces pushing it forward are equal and opposite to those pushing backwards.  We call this balanced forces.  If forces are unbalanced, then the object will accelerate until they do balance.

New bits  Kinetic energy = ½mV 2  M is mass, V is speed (velocity)  Notice, this is a square law, double the speed is four times the energy.

Stopping distances  Stopping a car or other vehicle takes time, during which it will travel a certain distance.  This stopping distance is made up of: Thinking distance – the distance it takes for the driver to react and start braking Braking distance – the distance it covers while the brakes and tyres stop the vehicle.

What affects……?  Thinking distance: Poor reactions – drink, drugs, tiredness, inexperience Failing to recognise hazards – inexperience, old age, poor visibility Distractions – noisy mates, ‘phone, stereo, smoking

What affects……?  Braking distance: Poor brakes – too little friction Worn or faulty tyres – too little friction Weather – ice, snow, rain – too little friction

What happens to stopping distance when you go faster?  Thinking distance increases in direct proportion to speed – double the speed, double the thinking distance, because it took the same time.  But the stopping distance is a square law  (Because your brakes take out the same amount of energy, but the energy is kinetic, so follows the law E = ½mV 2 )  Therefore double the speed gives four times the stopping distance.

From the Highway Code  See how doubling from 20mph to 40 gives braking distance going from 6m to 24m. (6 x 4 = 24)  Doubling from 30mph to 60 gives braking distance going from 14m to 55m. (14 x 4 = 56)  Brake effectiveness can vary depending on the speed (not part of this exam).