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PHYSICS UNIT 1: KINEMATICS (Describing Motion). MOTION ALONG A LINE. Who’s Upside Down?. MOTION ALONG A LINE. Who’s Moving?. MOTION ALONG A LINE. Motion : change in position of an object compared to a frame of reference (a "stationary" reference point) Measuring Motion (along a line)
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MOTION ALONG A LINE • Who’s Upside Down?
MOTION ALONG A LINE • Who’s Moving?
MOTION ALONG A LINE • Motion: change in position of an object compared toa frame of reference (a"stationary" reference point) • Measuring Motion (along a line) • Speed is the rate at which distance is covered. • Avg. speed = distance/time (m/s)
MOTION ALONG A LINE • Distance, D: Distance is how far an object moves, unit: m • Time, t: time since motion start, unit: s • Velocity, v: is speed in a given direction • Ex. 30 m/s North
MOTION ALONG A LINE • Instantaneous Speed: the speed at any given instant. (This is what your speedometer reads)
SOLVING PROBLEMS • Problem-Solving Strategy • Given: What information does the problem give me? S = 20 m/s t = 5 s • Question: What is the problem asking for? D = ? • Equation: What equations or principles can I use to find what’s required? S=D/t • Solve: Figure out the answer. D=Sxt = 100m • Check: Do the units work out correctly? Does the answer seem reasonable?
Practice • Two cars are traveling south on I-5. Car A has an average speed of 20.0 m/s. Car B has an average speed of 30.0 m/s. • a. How much time does it take Car A to travel 1500 m? • b. How far does Car B travel in 30.0 s? 75 s 900 m
LAB 1.1 QUIZ • A student made the following graph by moving in front of a motion sensor (the student was facing the sensor). Describe the student’s motion for each labeled section. b a e c d
PHYSICS UNIT 1: KINEMATICS (Describing Motion)
GRAPHING MOTION • interpreting an x vs. t (position vs. time) graph constant +v constant v = 0 constant –v changing +v changing +v (moving forward) (slowing down) (not moving) (moving backward) (speeding up)
x t GRAPHING MOTION • interpreting an x vs. t (position vs. time) graph • for linear x vs. t graphs: slope =rise/run =Dx/Dt, so rise = Dx slope = vav run = Dt
x t GRAPHING MOTION • interpreting an x vs. t (position vs. time) graph • for curving x vs. t graphs: slope of tangent line = v
GRAPHING MOTION • interpreting a v vs. t (velocity vs. time) graph constant +v constant v = 0 constant –v changing +v changing +v (slowing down) (moving backward) (speeding up) (not moving) (moving forward)
GRAPHING MOTION • comparing an x vs. t and a v vs. t graph
PHYSICS UNIT 1: KINEMATICS (Describing Motion)
constant velocity constant acceleration ACCELERATION
ACCELERATION • Acceleration, a: rate of change of velocity • unit: meter per second per second or m/s2 • speed increase (+a), speed decrease (–a), change in direction (what are the three accelerators in a car?) • average acceleration, aav = (vf-vi)/t= Dv/t
time (s) 0 1 2 3 4 5 6 speed (m/s) 0 2 4 6 8 10 12 position (m) 0 1 4 9 16 25 36 ACCELERATION • Constantacceleration example: a=2 m/s2 V = at D =1/2 a t2
terms: t: elapsed time xf : final position xi: initial position Dx: change in position (xf-xi) terms: a: acceleration vav: average velocity vf: final velocity vi: initial velocity Dv: change in velocity (vf-vi) ACCELERATION
defined equations: a = Dv/t vav = Dx/t vav = (vf+vi)/2 derived equations: Dx = ½(vf+vi)t vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2aDx ACCELERATION
QUIZ 1.3 • A train traveling 25.0 m/s begins slowing down with an acceleration of –4.00 m/s2. The train travels 50.0 m as it slows down (but it does NOT come to a stop).(a) What is its final velocity?(b) How much time does it take?(c) How far would the train have traveled in 5.00 s? Note: the train’s final velocity is NOT the same as it was in question (a) 15.0 m/s 2.50 s 75.0 m
PHYSICS UNIT 1: KINEMATICS (Describing Motion)
Free Fall: all falling objects are constantly accelerated due to gravity acceleration due to gravity, g, is the same for all objects use y instead of x, up is positive g = –9.80 m/s2(at sea level; decreases with altitude) GRAPHING MOTION
FREE FALL • air resistance reduces acceleration to zero over long falls; reach constant, "terminal" velocity
PHYSICS UNIT 1: KINEMATICS (Describing Motion)
Motion in a Plane vs. Motion in a Line It’s like reading a treasure map. Go 25 Paces North Go 15 paces West Go 30 paces North Go 20 paces Southeast X marks the Spot! MOTION IN A PLANE
MOTION IN A PLANE • Scalar Quantity: only shows how much size or magnitude (distance, time, speed, mass) • Vector Quantity: shows how much size or magnitude and in what direction • displacement, r : distance and direction • velocity, v : speed and direction • acceleration, a: change in speed and direction
q v N E W S MOTION IN A PLANE • Vectors • arrows:velocity vector v = v (speed), q(direction) • length proportional to amount • direction in map coordinates • between poles, give degreesN of W, degrees S ofW, etc.
Examples of Vectors • If a plane flies North at 100 m/s and the wind blows North at 20 m/s. What is the resultant? • If a plane flies North at 100 m/s and the wind blows South at 20 m/s then what is the resultant
MOTION IN A PLANE • Combining Vectors • draw a diagram & label the origin/axes! • Collinear vectors: v1 v2 v1 v2 • resultant: vnet=v1+v2 (direction: + or –) • ex: A plane flies 40 m/s E into a 10 m/s W headwind. What is the net velocity? • ex: A plane flies 40 m/s E with a 10 m/s E tailwind. What is the net velocity?
Combining non linear vectors • What if the plane were to fly into a cross wind? • Ex. A plane flies North at 100 m/s and there is a 50 m/s Easterly wind. What is the plane’s net (combined) velocity?
MOTION IN A PLANE • Perpendicular vectors: resultant’s magnitude: resultant’s direction:
PHYSICS UNIT 1: KINEMATICS (Describing Motion)
UNIT 1 TEST PREVIEW • Concepts Covered: • motion, position, time • speed (average, instantaneous) • x vs. t graphs, v vs. t graphs, a vs. t graphs • vectors, scalars, displacement, velocity • adding collinear & perpendicular vectors • acceleration • free fall, air resistance
UNIT 1 TEST PREVIEW • What’s On The Test: • 21 multiple choice, 12 problems Dx = ½(vf+vi)t vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2aDx