Average velocity is displacement divided by the time interval.
The total displacement divided by the time interval during which the displacement occurred (Vavg) Basically velocity is distance/time
Average Velocity The units can be determined from the equation. SI Units: m/s Other Possible Units: mi/h, km/h, cm/year Velocity can be positive or negative but time must always be positive
Average Velocity You travel 370 km west to a friends house. You left at 10am and arrived at 3pm. What was your average velocity? Vavg = d = -370 km Δt 5.0h Vavg= -74 km/h or 74 km/h west
Classroom Practice Problems Joey rides his bike for 15 min with an average velocity of 12.5 km/h, how far did he ride? Vavg = d Δt d = (12.5 km/h) (0.25h) d = 3.125 km
Speed Speed does not include direction while velocity does. Speed uses distance rather than displacement. In a round trip, the average velocity is zero but the average speed is not zero.
Graphing Motion How would you describe the motion shown by this graph? Answer: Constant speed (straight line) What is the slope of this line? Answer: 1 m/s What is the average velocity? Answer: 1 m/s
Graphing Motion Describe the motion of each object. Answers Object 1: constant velocity to the right or upward Object 2: constant velocity of zero (at rest) Object 3: constant velocity to the left or downward
What do you think? Which of the following cars is accelerating? A car shortly after a stoplight turns green A car approaching a red light A car with the cruise control set at 80 km/h A car turning a curve at a constant speed Based on your answers, what is your definition of acceleration?
Acceleration What are the units? SI Units: m/s2 Other Units: (km/h)/s or (mi/h)/s Acceleration = 0 implies a constant velocity (or rest)
Classroom Practice Problem A bus slows down with an average acceleration of -1.8m/s2. how long does it take the bus to slow from 9.0m/s to a complete stop? Find the acceleration of an amusement park ride that falls from rest to a velocity of 28 m/s downward in 3.0 s. 9.3 m/s2 downward
Acceleration Displacement with constant acceleration d = ½ (vi + vf) Δt Displacement= ½ (initial velocity + final velocity) (time)
Example A racecar reaches a speed of 42m/s. it uses its parachute and breaks to stop 5.5s later. Find the distance that the car travels during breaking. Given: vi = 42 m/s vf= 0 m/s t = 5.5 sec d = ??
Example d = ½ (vi + vf) Δt d = ½ (42+0) (5.5) d = 115.5 m
Acceleration Final velocity Depends on initial velocity, acceleration and time Velocity with constant acceleration vf = vi + aΔt Final velocity = initial velocity + (acceleration X time)
Acceleration Displacement with constant acceleration d = viΔt + ½ a (Δt)2 Displacement = (initial velocity X time) + ½ acceleration X time2
Example A plane starting at rest at one end of a runway undergoes uniform acceleration of 4.8 m/s2 for 15s before takeoff. What is its speed at takeoff? How long must the runway before the plane to be able to take off? Given: vi = 0 m/s a= 4.8m/ss t= 15s vf = ?? d= ??
Example What do we need to figure out? Speed at takeoff How long the runway needs to be vf = vi + aΔt vf = 0m/s + (4.8 m/s2)(15 s) vf = 72m/s
Example d = viΔt + ½ a (Δt)2 d = (0m/s)(15s) + ½ (4.8m/s2)(15s)2 d = 540 m
Acceleration Final velocity after any displacement vf2 = vi2 +2ad Final velocity2 = initial velocity2 + 2(acceleration)(displacement)
Example An aircraft has a landing speed of 83.9 m/s. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration required for safe landing? Answer: -18.0 m/s2
Useful Equations 1. 2. 3. 4. 5. d = 1 (vi + vf) t 2
Classroom Practice Problems A bicyclist accelerates from 5.0 m/s to 16 m/s in 8.0 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval? Answer: 84 m
Free Fall The motion of a body when only the force due to gravity is acting Acceleration is constant for the entire fall Acceleration due to gravity (ag or g) Has a value of -9.81 m/s2 (we will use 10m/s2) Negative for downward Roughly equivalent to -22 (mi/h)/s
Free Fall Acceleration is constant during upward and downward motion Objects thrown into the air have a downward acceleration as soon as they are released.
Free Fall The instant the velocity of the ball is equal to 0 m/s is the instant the ball reaches the peak of its upward motion and is about to begin moving downward. REMEMBER!!! Although the velocity is 0m/s the acceleration is still equal to 10 m/s2
Example Jason hits a volleyball so that it moves with an initial velocity of 6m/s straight upward. If the volleyball starts from 2.0m above the floor how long will it be in the air before it strikes the floor? We want to use our velocity and acceleration equations
Example Given: vi = 6m/s a= 10m/s2 d= -2.0m t = ?? vf = ?? vf2= vi2 + 2ad vf= at + vi
Free Fall For a ball tossed upward, make predictions for the sign of the velocity and acceleration to complete the chart.
Graphing Free Fall Based on your present understanding of free fall, sketch a velocity-time graph for a ball that is tossed upward (assuming no air resistance). Is it a straight line? If so, what is the slope? Compare your predictions to the graph to the right.
Classroom Practice Problem A ball is thrown straight up into the air at an initial velocity of 25.0 m/s upward. Create a table showing the ball’s position, velocity and acceleration each second for the first 5 s.
Direction of Acceleration Describe the motion of an object with vi and a as shown to the left. Moving right as it speeds up Moving right as it slows down Moving left as it speeds up Moving left as it slows down
Graphing Velocity The slope (rise/run) of a velocity/time graph is the acceleration. Rise is change in v Run is change in t This graph shows a constant acceleration. Average speed is the midpoint.
Graph of v vs. t for a train Describe the motion at points A, B, and C. Answers A: accelerating (increasing velocity/slope) to the right B: constant velocity to the right C: negative acceleration (decreasing velocity/slope) and still moving to the right