Distance, Speed and Time speed (m/s) = distance (m) time (s) x D S T S = D T D = S X T T = D S

1.An athlete runs 200m in 20s. What is his speed? speed = distance time speed = 200 m 20 s speed = 10 m/s

2.A bus travels 2000m between stops at a speed of 40m/s. Find the time taken to travel this distance. speed = distance time 40 = 2000 time time = = 50s

3. It takes a cyclist 30s to go 850 m. What is her speed? speed = distance time speed = 850 m 30 s speed = m/s

4.A walker travels at 3m/s for a distance of 50m. How long did it take him? speed = distance time 3 = 50 time time = 50 3 = s

This car is travelling at a velocity of 20m/s east This car is travelling at a speed of 20m/s VELOCITY AND SPEED

x Distance and Displacement Distance is simply how far something has moved Man and dog have walked 3 metres from X x Man and dog have walked 3 metres WEST from X Displacement is ‘distance in a given direction’

Acceleration acceleration (m/s 2 ) = change in velocity (m/s) time taken (s) Remember: the formula for acceleration and deceleration is the same

What does acceleration mean? The rate at which speed increases e.g. 5 m/s 2 5m/s/s The speed is increasing by 5m/s every second 0s 1s 2s 3s 4s 5s 6s 5m/s 10m/s 15m/s 20m/s 25m/s 30m/s

Questions on acceleration 1.A cyclist accelerated from rest to 10m/s in 5s. Find the acceleration. acceleration = change in velocity time = 10 – 0 = = 2 m/s 2

2.A sprinter crosses the finishing line at a speed of 8m/s and comes to rest 10s later. Calculate the deceleration of the sprinter. acceleration = change in velocity time = 8 – 0 = = 0.8 m/s 2

3.A grand prix car increases its speed from 50m/s to 120m/s in 3s. What is the acceleration? acceleration = change in velocity time = 120 – 50 = = m/s 2

4. A skier increases his speed from 6m/s to 10m/s in 2s. What is his acceleration? acceleration = change in velocity time = = = 2 m/s 2

MASS Mass is the amount of matter inside an object. It is measured in kilograms (kg) 15 kg 50 kg

GRAVITATIONAL FIELD STRENGTH On Earth = 10N/kg g = 10

To find the WEIGHT of of an object you need to know the relationship between WEIGHT, MASS and GRAVITATIONAL FIELD STRENGTH weight (N) = mass (kg) x gravitational field strength (N/kg) 1kg g = N = 1 x 10 = 10 N 1kg

Weight is the force of gravity on a body and is measured in NEWTONS (N). On Earth the force on a 1kg object is 10N (g = 10N/kg). On the moon the force on a 1kg object is 1.67N (g = 1.67N/kg) 50kg What is the weight of the Martian if it lands on the Earth or the Moon?

W GM To find the weight of the Martian you need to know:- weight (N) = mass (kg) x gravitational field strength (N/kg) W = M X G M = W G G = W M

Earth (g = 10) weight = mass x g = 50 x 10 = 500 N 50 kg

Moon ( g = 1.67) 50 kg Weight = mass x g = 50 x 1.67 = 83.5 N

Thinking distance Braking distance STOPPING DISTANCE OF A VEHICLE OVERALL STOPPING DISTANCE

Stopping a vehicle Braking distance Alcohol Thinking distance Tiredness Drugs Wet roads Driving too fast Tyres/brakes worn out Icy roads Visibility

Stopping a vehicle Condition of road wet/icy roads Condition of vehicle Condition of driver alcohol/drugs concentration type of road surface brakes tyres

Distance-Time Graphs Distance (metres) Time/s Constant Speed FASTEST Constant Speed Constant Speed At Rest

Distance-time graphs (outward/return journey) Distance (metres) Time/s outward journey return journey

Distance-time graphs (total journey time/total distance travelled) Distance (metres) Time/s 80m 100s 40m

Distance-time graphs (finding speed from part of graph) Distance (metres) Time/s speed during 1 st 20s = distance = 10m = 0.5m/s time 20s Distance-time graphs (finding speed from part of graph)

Distance-time graphs (average speed for journey) Distance (metres) Time/s average speed = 80m = 0.8 m/s 100s Distance-time graphs (average speed for journey)

m 200m 300m 400m 500m Time (s) distancedistance 0 DISTANCE / TIME GRAPH A B C D E (m)

Questions on Distance/Time Graph Find: 1.the average speed during A-B S = D = 300m = 15m/s T 20s 2.the average speed between B – C S = D = 0m = 0m/s T 20s 3. the average speed for the return journey S = D = 460m = 7.67 m/s T 60s

Speed/Time graphs Speed (m/s) Time (s) This object is travelling at a constant speed of 4m/s for 5s

Speed/time graphs (acceleration - increasing speed) speedspeed (m/s) time (s) This object is increasing speed from 0m/s (rest) to 10m/s in 5s

Speed/time graphs (finding acceleration) speedspeed (m/s) time (s) acceleration = change in velocity time = 10 – 0 5 = 2 m/s 2

speed/time graphs (average speed) speedspeed (m/s) time (s) Average speed = = 5 m/s 5 m/s

speed/time graphs (deceleration -decreasing speed) speedspeed (m/s) time (s) This object is decreasing its speed from 20m/s to 0m/s (rest) in 5s

speed/time graphs (finding deceleration) speedspeed (m/s) time (s) deceleration = change in velocity time = 20 – 0 5 = 4 m/s 2

speed/time graphs (average speed) speedspeed (m/s) time (s) Average speed = = 10 m/s 10 m/s

Speed/Time graph TIME (S) VELOCITY (M/S) accelerating decelerating constant speed

What is the acceleration in the 1 st 20s? Acceleration = change in velocity time = 30 – 0 20 = 1.5m/s 2

What is the deceleration in the last 10s? deceleration = change in velocity time = 40 – 0 10 = 4m/s 2

Speed/Time graphs (finding distance travelled) Speed (m/s) Time (s) distance = area under the graph distance = speed x time = height x length = 4 x 5 = 20m area of the green rectangle

speed/time graphs (distance travelled) Area under graph = b x h 2 = 5 x 10 2 = 25 m speedspeed (m/s) time (s)

speed/time graphs (distance travelled) speedspeed (m/s) time (s) Distance travelled = b x h 2 = 5 x 20 2 = 50 m

Speed/Time graph (finding distance travelled) TIME (S) VELOCITY (M/S)

What distance is travelled between 0s & 20s? Area = b x h 2 = 20 x 40 2 = = 400m 0

What distance is travelled between 20s & 50s? Area = b x h = 40 x 30 0 = 1200m

What distance is travelled between 50s & 60s? = 10 x = 200m Area = b x h 2

Speed/Time graph (bigger the area under the graph ………..) TIME (S) VELOCITY (M/S) Bigger the area under the graph the further the distance travelled

RESULTANT FORCES When 2 or more forces act on an object they can be replaced by a single force called the RESULTANT FORCE 6N 12N

RESULTANT FORCES 4N 0N

RESULTANT FORCES 29N 16N 1N 12N

BALANCED FORCES 3N OBJECT AT REST (STATIONARY) RESULTANT FORCE = 3N – 3N = 0N OBJECT REMAINS AT REST (STATIONARY)

BALANCED FORCES 3N MOVING OBJECT RESULTANT FORCE = 3N – 3N = 0N OBJECT TRAVELS AT CONSTANT SPEED

UNBALANCED FORCES 6N 3N OBJECT AT REST (STATIONARY) RESULTANT FORCE = 6N – 3N = 3N OBJECT INCREASES SPEED FROM REST (ACCELERATES)

UNBALANCED FORCES 6N 3N MOVING OBJECT RESULTANT FORCE = 6N – 3N = 3N OBJECT INCREASES SPEED FROM SPEED ORIGINALLY MOVING AT

UNBALANCED FORCES 3N 6N MOVING OBJECT RESULTANT FORCE = 6N – 3N = 3N OBJECT SLOWS DOWN FROM SPEED ORIGINALLY MOVING AT (DECELERATES)

BALANCED FORCES FALLING OBJECT WEIGHT/GRAVITY AIR FRICTION/RESISTANCE FALLS AT A CONSTANT SPEED (TERMINAL VELOCITY)

UNBALANCED FORCES FALLING OBJECT WEIGHT/GRAVITY AIR FRICTION/RESISTANCE OBJECT INCREASES SPEED (ACCELERATES) 2N 6N

UNBALANCED FORCES FALLING OBJECT WEIGHT/GRAVITY AIR FRICTION/RESISTANCE OBJECT DECREASES SPEED (DECELERATES) 8N 6N REMEMBER: THE WEIGHT OF THE OBJECT REMAINS THE SAME NO MATTER WHAT SPEED IT FALLS AT

UNBALANCED FORCES WEIGHT/GRAVITY AIR FRICTION/RESISTANCE 2N 6N AS SPEED INCREASES THE RESISTIVE FORCES INCREASE 5N 6N 20m/s 30m/s WEIGHT/ GRAVITY ALWAYS STAYS THE SAME

ACTION/REACTION FORCES For every ACTION force there is an EQUAL but OPPOSITE REACTION force

Wall pushing man Man pushing wall

BULLET RECOIL REACTION FORCE ACTION FORCE

What 2 things affect the acceleration of this car? MASS measured in kilograms RESULTANT FORCE, the difference between the force due to friction and the force of the engine

Resultant Force, Mass and Acceleration resultant force (N) = mass (kg) x acceleration (m/s 2 ) F AM f = m x a a = f m m = f a

motion 10N f = m x a 0 = 5 x a a = 0 = 0m/s 2 = constant speed 5 5kg

motion 20N15N f = m x a 5 = 3 x a a = 5 = 1.67 m/s 2 3 3kg

55m/s 2 300N f = m x a 300 = m x 55 m = 300 = 5.45 kg 55

ACCELERATION DUE TO GRAVITY If an elephant and a feather were dropped from the same height, which one would hit the ground first?

Due to air resistance on the feather and the elephants greater mass the elephant would hit the ground first

If we ignore air resistance the elephant and feather would land on the ground at the same time due to their ACCELERATION DUE TO GRAVITY being the same!!! All objects will accelerate downwards at the same rate if air friction is ignored so they will hit the ground at the same time! The acceleration g = 10m/s 2