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SCALARS 35 km long 35 mm wide 3 hours 35 minutes 35.8 kg 2000 kcal 20% efficiency Vectors 5 km south 35 m/s south-west 45 N upward 9.8 m/s2 downward Symbol for vectors: Typed: v, a, F, x Written: v, a, F, x Multiplying vectors by scalars Dx = v × Dt Vector x scalar = vector Multiplying vectors by scalars a 2a -2a 1 a 2 Comparing vectors Which vectors have the same magnitude? Which vectors have the same direction? Which arrows, if any, represent the same vector? Adding vectors a b a+b a+b b+a b+a Subtracting vectors (+ negative) a a-b a-b b b-a b-a Subtracting vectors (“fork”) a a-b b a-b b-a b-a Check your understanding Construct and label a diagram that shows the vector sum 2A + B. Construct and label a second diagram that shows B + 2A. Construct and label a diagram that shows the vector sum A – B/2. Construct and label a second diagram that shows B/2 - A. Adding vectors algebraically Adding vectors that are perpendicular Example A person walks 5 km east and 3 km south. What is the person’s total displacement? Hint: displacement is a vector quantity, the answer should have magnitude (km) and direction (degrees) 1km Dx = Dx1 + Dx2 Dx = x + x 2 1 2 2 Dx = 25 + 9 æ 3ö q = tan ç ÷ è 5ø -1 Relative Motion • A person is sitting in a bus that is moving north 2 m/s. a)What is the passenger’s velocity relative to a person standing on the sidewalk nearby? b) What is the passenger’s velocity relative to the person sitting next to him? c) What is his speed relative to a car moving at 4 m/s in the opposite late towards the bus? Relative motion • A swimmer moves at 2 m/s in a pool. a)What will be her speed relative to a person on the river bank if she swims downstream and the current is 5 m/s? b) What will her speed be if she tries to swim against the current? • A boat is crossing a river moving south with a speed of 4.0m/s relative to the current. The current moves due east at a speed of 2.0 m/s relative to the land. a) How far across the river will the boat be in 2 seconds? b) How far down the river the boat be in 2 seconds? c) What will the boat’s total displacement be (from the place where it started). Relative motion • A motorboat is crossing a 350-m wide river moving 10 m/s relative to water. If current is 3 m/s east, how far down the stream will the boat end up when it reaches the opposite bank? What will the boat’s total displacement be? What will the boat’s velocity be relative to water? • (more than one way to solve!) Adding vectors that are NOT perpendicular Adding vectors algebraically Adding vectors Adding vectors RESOLVING VECTORS b a yacomponent y q x component a x ax = a × cosq ay = a ×sinq ax = a ×sin b ay = a × cos b NE EN How fast must a truck travel to stay beneath an airplane that is moving 105 km/h at an angle of 25° to the ground? vt = v p cos 25 vt = 105×cos25 = 95(km / h) Practice finding components • • • • 35 km/h 20o NE 120 m 35o WN 3.2 m/s2 40o SW 65 km 70o ES X: 33 km/h Y: 12 km/h X: - 69 m Y: 99 m X: - 2.5 m/s2 Y: - 2.1 m/s2 X: 61 km Y: - 22km 40 20 70 35 In groups, on big graph paper, illustrate the following • One of the holes on a golf course lies due east of the tee. A novice golfer flubs his tee shot so that the ball lands only 64 m directly northeast of the tee. He then slices the ball 30° south of east so that the ball lands in a sand trap 127 m away. Frustrated, the golfer then blasts the ball out of the sand trap, and the ball lands at a point 73 m away at an angle 27° north of east. At this point, the ball is on the putting green and 14.89 m due north of the hole. To his amazement, the golfer then sinks the ball with a single shot. • Use algebraic formulas to find the x and y components of each displacement vector. Shot 1 x component _____________ y component _____________ Shot 2 x component _____________ y component _____________ Shot 3 x component _____________ y component _____________ Shot 4 x component _____________ y component _____________ Calculate • Find the total displacement (to the nearest meter) the golf ball traveled from the tee to the hole. Assume the golf course is flat. (Hint: Which component of each displacement vector contributes to the total displacement of the ball between the tee and the hole?) Adding vectors Adding vectors algebraically R= R +R 2 x 2 y Rx = ax + bx Ry = ay + by æ Rx ö q = tan ç ÷ or è Ry ø -1 æ Ry ö q = tan ç ÷ è Rx ø -1 Example, p. 94, #1 • A football player runs directly down the field for 35 m before turning to the right at an angle of 25° from his original direction and running an additional 15 m before getting tackled. What is the magnitude and direction of the runner’s total displacement? ax = 0m; ay = 35m bx = 15sin 25 = 6.34m b = 15m by = 15 cos 25 = 13.6m Rx = 0 + 6.34 = 6.3m Ry = 35 +13.6 = 49m a = 35m R = R2x + R2y = 6.32 + 49 2 = 49(m) æ 6.3 ö q = tan ç ÷ = 7.3 è 49 ø -1 Example 2, p 94, #4 • An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 75 km 30.0° west of north, and the second is 155 km 60.0° east of north. What is the total displacement of the airplane? Answer: 171 km at 34° east of north Table hockey lab Projectile motion Free-falling while moving horizontally Moving in two directions at the same time Projectile motion Parabolic trajectory Vertical motion: with acceleration of -9.8 m/s2 and initial velocity vy and vertical displacement Δy Horizontal motion: constant vx and horizontal displacement Δx Projectile motion g Vertical Horizontal Accelerated Constant speed v t 9.8 m / s 2 v v f vi g t y vi t 1 t vi v f v g (t ) 2 x vt 2 v f vi 2 g y 2 x 2 v vertical com ponen t t x v v horizontal com ponent An object launched horizontally Find the initial speed of the bike Given Dy = -50.0m g = -9.81m / s 2 vyi = 0 Dx = 90.0m vx = ? Equations / Substitution Horizontal Vertical Dx vx = Dt 1 Dy = viy Dt + g(Dt)2 2 1 Dy = g(Dt)2 2 2Dy Dt = g Solving vx = 90 = 3.2(s) 3.2 2 ×(-50) -9.81 Dt = 3.19(s) 90.0 vix = 3.19 Dt = vix = 28.2m / s Lab – Launching a Projectile Horizontally • Work in pairs • Each person should have his / her set of data • Each student turns in a report with his / her name on it. Projectile Launched at an angle Projectile has initial velocity in both directions vxi = vcos 30 vx = const vyi = vsin 30 vyf = vyi + gDt Example • A football is kicked 22 m/s at 40o to the horizontal. Calculate the maximum height and horizontal range the football will reach before it falls down (providing no one catches it). 1st step: v ® vx and v y v = 22m/s at q =40 v y = 22 ×sin 40 vx = 22 ×cos 40 H y v v 2 g y ( v f 0) 2 f 2 i y v 2 i g t t x vt t ? 2g v R x 0 14 9.8 t 2 1.44 2.9( s ) R 17 2.8 49( m ) Example 2 (p.110, #35) • A place kicker must kick a football from a point 36.0 m from the goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53° to the horizontal. • a. By how much does the ball clear or fall short of clearing the crossbar? • b. Does the ball approach the crossbar while still rising or while falling? Given vi = 20m / s Units Equations vx = 20×cos53 v y = 20×sin53 q = 53 hc = hg - Dy hg = 3.05m Dy = ? vy ¹ 0 hc = ? 1 Dy = viy Dt + g(Dt)2 2 - or ¯ Dt = Dx vx - or ¯ Dt > or < Dtup 0 - v yi Dtup = g Substitute/Solve Dt = 36 = 2.99(s) 1.204 1 Dy = 16 × 2.99 - 9.8×(2.99)2 2 Dy = 4.03m hc = 4.03- 3.05 = .98(m) Dtup = 0 -16 = 1.63(s) -9.8 Dt > Dtup Þ ¯