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## Must Know.....

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**Must Know.....**Your Subtitle Goes Here**ti = initial time**• tf = final time • ∆t = time interval = tf - ti • xi = initial position • xf = final position • ∆x = displacement = xf– xi • distance = s∙∆t • s = speed**Motion Diagram**• Motion diagrams are a pictorial description of an object in motion. (Ex)**Changing Velocity (Acceleration)**• Speeding up = going faster and faster = acceleration (a) Velocity increases at a constant rate (Ex) 10 m/s, 15 m/s, 20 m/s, 25 m/s, 30 m/s (b) Velocity increases at a decreasing rate (Ex) 10 m/s, 15 m/s, 19 m/s, 22 m/s, 24 m/s, 25 m/s 2. Slowing down = going slower and slower = deceleration (Ex) 10 m/s, 8 m/s, 6 m/s, 4 m/s, 2 m/s 3. Changing the direction (Ex) 10 m/s, -10 m/s Caution: Outside the classroom, acceleration means to speed up**Constant Acceleration**• The velocity increases at a constant rate (Ex) A car’s velocity at 1:00 pm is 0 mph; at 1:01 pm, 20 mph; at 1:02 pm, 40 mph; 1:03 pm, 60 mph; 1:04 pm, 80 mph • The velocity decreases at a constant rate (Ex) A car’s velocity at 1:05 pm, 80 mph; at 1:06 pm, 70 mph; at 1:07 pm, 60 mph; 1:08 pm, 50 mph; 1:09 pm, 40 mph**You must be able to tell the velocity is changing...**• from the motion diagram • from the velocity vs. time graph 3. from the table of: (a) velocity vs. time (b) position vs. time*** Cars with leaky oil – PDF file..\Physics Labs &**Worksheets • Is this what you got? ..\Cars with leaking oil.xlsx**Average Acceleration**• a vector quantity (Ex) A car’s velocity at 11:00 AM was 90 km/hr and its velocity changed to 110 km/hr at 11:03 AM. What is the average acceleration of this car?**Which shows acceleration?**• displacement vs. time**velocity and acceleration**• Determine the acceleration in each motion diagram. vi = 5 m/s vf = 7 m/s vi = 5 m/s vf = 3 m/s vf = -7 m/s vi = -5 m/s vf = -3 m/s vi = -5 m/s**More on Motion with Constant Acceleration**velocity, m/s m = ā vi time, s**Constant (Average) Acceleration**Quick Review: For an object moving at a constant velocity,**Relating Velocity & Acceleration**• “+” velocity and “+” acceleration = Demo • “+” velocity and “–” acceleration = Demo • “‒” velocity and “+” acceleration = Demo • “‒” velocity and “‒” acceleration = Demo**Position, Velocity & Acceleration**acceleration, m/s2 0 time, s**Practices**velocity, m/s A B E D 0 time, s C**Examples**• A train moving with a velocity of 51 m/s east undergoes an acceleration of -2.3 m/s2 as it approaches a town. What is the velocity of the train 5.2 s after it has begun to decelerate? • After being launched, a rocket attains a speed of 122 m/s before the fuel in the motor is completely used. If you assume that the acceleration of the rocket is constant at 32.2 m/s2, how much time does it take for the fuel to be completely consumed?**3. A car traveling at 21 m/s misses the turnoff on the road**and collides into the safety guard rail. The car comes to a complete stop in 0.55 s. a) What is the average acceleration of the car? b) If the safety rail consisted of a section of rigid rail. the car would stop in 0.15 s. What would be the acceleration in this case? 4. A cheetah can reach a top speed of 27.8 m/s in 5.2 s. What is cheetah’s average acceleration?**Position, Velocity & Acceleration**• Given: • (position or displacement missing) • (final velocity missing) • (time missing) • In all of above equations, ∆t=tf if ti= 0 • See Pg 52-56 for the derivation**Words to Formulas**• decide the positive direction • to right, to east, or up: positive • to left, to west, or down: negative • come to stop or to rest: vf= 0 • from rest: vi = 0 • slows down: +v and ‒a, or ‒v and +a • speeds up: +v and +a, or ‒v and ‒a • at constant velocity: a = 0 • going down or up a hill: a ≠ 0 • the sign of ∆v = the sign of ∆d • tf = ∆t = a positive number (ti = 0) • df= ∆d (di = 0) • at the moment of reversing direction, v = 0 and a ≠ 0 • how long to take: ∆t • how far: ∆d • from A to B: A = initial; B = final**Examples**• Suppose a car rolls down a 52.0-m long inclined parking lot and is stopped by a fence. If it took the car 11.25 s to roll down the hill, what was the acceleration of the car before striking the fence? • A sky diver in free fall reaches a speed of 65.2 m/s when she opens her parachute. The parachute quickly slows her down to 7.30 m/s at a constant rate of 29.4 m/s2. During this period of acceleration, how far does she fall?**3. A child rolls a ball up a hill at 3.24 m/s. If the ball**experiences an acceleration of 2.32 m/s2, how long will it take for the ball to have a velocity of 1.23 m/s down the hill? 4. A cheetah can accelerate from rest to a speed of 27.8 m/s in 5.20 s. The cheetah can maintain this speed for 9.70 s before it quickly runs out of energy and stops. What distance does the cheetah cover during this 14.9-s run?