Movement and Position, Speed and velocity.ppt
- 1. Average Speed average speed = distance time In physics speed is usually measured in: metres per second (m/s)
- 2. also: distance = and: time = speed time distance speed x time distance speed
- 3. Speed Conversions 1 kilometre per hour (km/h) = 1000 metres per hour but 1 hour = 3600 seconds therefore 1 km/h = 1000m ÷ 3600 s 1 km/h = 0.28 m/s and 1 m/s = 3.6 km/h Also: 100 km/h = approx 63 m.p.h
- 4. Question 1 Calculate the average speed of a car that covers 500m in 20s. average speed = distance time = 500m / 20s = 25 m/s (about 60 mph)
- 5. Question 2 Sound waves travel at about 340m/s through air. How far will a sound wave travel in one minute? distance = speed x time = 340 m/s x 1 minute = 340 m/s x 60 seconds = 20 400 m (20.4 km or about 13 miles)
- 7. The slope or gradient of a distance-time graph increases with speed. distance time slow f a s t
- 8. The slope or gradient of a distance-time graph is equal to the speed. In the graph opposite: slope = 150m / 10s = 15 m/s = speed
- 9. Choose appropriate words to fill in the gaps below: Speed is equal to ________ divided by time and can be measured in _________ per second. A speed of 20 m/s is the same as ______ km/h which is approximately equal to ______ mph. The _________ of a distance against time graph can be used to calculate ________. The greater the gradient of the line the __________ is the speed. The line will be ___________ when the speed is zero. higher 72 40 distance slope metres WORD SELECTION: speed horizontal higher 72 40 distance slope metres speed horizontal
- 10. Velocity The velocity of a body is its speed in a given direction. The airplane opposite may loop at a constant speed but its velocity changes as its direction of motion changes.
- 11. Question A stone dropped off the top of a cliff falls down by 20m in 2s. Calculate its average velocity (a) downwards and (b) horizontally. (a) average speed downwards = 20m / 2s = 10m/s Therefore velocity downwards = 10 m/s (b) average speed horizontally = 0m / 2s = 0m/s Therefore velocity horizontally = 0 m/s
- 12. Acceleration acceleration = velocity change time taken a = (v – u) / t a = acceleration in metres per second squared (m/s2 ) v = final velocity in m/s u = initial velocity in m/s t = time taken in seconds (s)
- 13. Why is acceleration measured in m/s2 ? acceleration = velocity change time taken velocity change is measured in m/s time taken is measured in s therefore acceleration = m/s ÷ s = m/s2
- 14. Other notes: 1. Speed and velocity: Often, but not always, speed can be used in the equation. 2. Change in velocity: = final velocity – initial velocity = v - u 3. Deceleration: This is where the speed is decreasing with time. 4. Circular motion at a constant speed: Acceleration is occurring because the direction of motion is continually changing and hence so is velocity.
- 15. Question 1 Complete the table below for an airplane accelerating at 8m/s2 . time (s) 0 1 2 3 4 velocity (m/s) 0 8 16 24 32
- 16. Question 2 Calculate the acceleration of a car that changes in velocity from 5m/s to 25m/s in 4 seconds. a = (v – u) / t = (25m/s – 5m/s) / 4s = 20 / 4 acceleration = 5 m/s2
- 17. Question 3 Calculate the final velocity of a train that accelerates at 0.3m/s2 for 60 seconds from an initial velocity of 5m/s. a = (v – u) / t becomes: (v – u) = a x t = 0.3m/s2 x 60s = 18m/s therefore final train velocity = 5m/s + 18m/s = 23 m/s
- 18. Question 4 Calculate the deceleration of a car that slows down from 18m/s to rest in 3 seconds. a = (v – u) / t = (0m/s – 18m/s) / 3s = -18 / 3 (notice minus sign) acceleration = - 6 m/s2 and so deceleration = 6 m/s2 Note: Deceleration is the negative of acceleration.
- 19. Choose appropriate words to fill in the gaps below: Velocity is speed measured in a particular ______________. A person walking northwards will have _______ velocity in a westwards direction. Acceleration is equal to ________ change divided by the time taken. Acceleration is measured in metres per second ______. Deceleration occurs when a body is _________ down. It is possible for a body to be accelerating even when its ______ is not changing provided its direction is, for example: a body moving in a ________. slowing circle zero speed direction squared velocity WORD SELECTION: slowing circle zero speed direction squared velocity
- 20. Velocity-time graphs velocity time low acceleration h i g h a c c e l e r a t i o n The slope of a velocity-time graph represents acceleration. deceleration constant velocity or zero acceleration
- 21. area equals distance travelled The area under a velocity-time graph represents distance travelled. velocity time
- 22. Question 1 Sketch the velocity time graph of a car accelerating from rest to 15m/s in 3 seconds and then remaining at a constant speed for one more second. velocity (m/s) time (s) 1 2 3 4 15 5 10
- 23. area Question 2 Calculate the acceleration and the distance travelled after 4 seconds from using the graph opposite. velocity (m/s) time (s) 1 2 3 4 12 4 8 acceleration = gradient = y-step ÷ x-step = (12 - 0)m/s ÷ (4 – 0)s = 12 / 4 acceleration = 3 m/s2 distance = area under the graph = area of triangle = ½ x base x height = ½ x 4s x 12m/s distance travelled = 24m
- 24. Question 3 Calculate the acceleration and distance travelled using the graph shown below.
- 25. Acceleration: Acceleration equals the slope of the graph = y-step ÷ x-step = (12 - 4)m/s ÷ (10s) = 08 / 10 Acceleration = 0.8 m/s2 Distance travelled: This equals the area below the graph = area of rectangle + area of triangle = (10s x 4m/s) + (½ x 10s x (12 – 4)m/s) = 40m + 40m Distance travelled = 80m
- 26. Question 4 Calculate the distance travelled over 15 seconds and the deceleration during the final five seconds using the graph below.
- 27. Distance travelled: This equals the area below the graph = area of rectangle + area of triangle = (10s x 20m/s) + (½ x 5s x 20m/s) = 200m + 50m Distance travelled = 250m Deceleration: Acceleration equals the slope of the graph = y-step ÷ x-step = (- 20m/s) ÷ (5s) = - 4 m/s2 but deceleration = negative of acceleration Deceleration = 4 m/s2