1.1 Speed and velocity in Motion in One and Two Dimensions kinematics: the study of motion, the study of moving objects

1-D Motion One dimensional motion or linear motion occurs when motion can only occur in two directions which are opposite to each other. (One set of directions) Example: A train on a straight track can only move forwards or backwards.

2-D Motion Two dimensional motion is motion in a plane. (Two sets of directions) Example: A person running all over a level field; the person can go east, west, south, and north.

Scalar Quantities Scalar quantities have a number (also called magnitude or size) and unit, but no direction. Examples: distance, speed, time, energy, and power

Vector Quantities Vector quantities have number (also called magnitude or size), unit, and direction. Examples: position, displacement, velocity, acceleration, and force

Defining Direction Mathematically Direction does not fit into an equation well We must describe the direction of motion using mathematical language To do this we define direction as either + or – This should be done at the beginning of each problem N + W - E + S -

If you always refer to the horizontal… Vertical component Magnitude sin(angle) Horizontal component Magnitude cos (angle) Add the sign and then you are done!

Distance describes the length of the path travelled (scalar quantity) Displacement describes a change in position (includes direction) → Its symbol is Δd and its SI unit is metres, m. → → → → → → Δd = d2 – d1 or Δd = df – di Note that d represents position

Displacement and velocity in Two Dimensions: The total distance is simply the addition of the magnitude of each displacement vector. The best way to solve for 2D resultant displacement is to break each vector into its components.

1.1 Speed and velocity Note The best problem solving method is D-GRASS. D: Diagram, draw the situation if possible or if you can’t visualize it in your head G: Given information, the values in the problem R: Required, what you are asked to find

1.1 Speed and velocity Note A: Analysis, determine what information can be used with equations S: Solve, substitute into equations and calculate the desired value S: Sentence, Write a statement including the required value as a proper sentence

Sample Problems – Right Angle Christine walks 15 m [E] and 10 m [N]. What is her displacement?

Sample Problem – Non-Right Angle Sifat walks 15 m [E] and then 10 m [E10S]. What is his displacement?

Sample Problem - Components What is the resultant displacement of an ATV that goes 4.0 km [N], 5.0 km [E], 7.5 km [SE], 11.0 km [20º N of E], and finally 8.8 km [30º E from N].

D-Grass outline for Example #2 1. Make a diagram (Include individual displacements) 2. Required; resultant displacement = ? 3. Analysis…Break into component, total each type of component (x and y), and combine them.

D-Grass outline for Example #2 4. Solve to find the resulting total displacement for each component (x and y). And then calculate the magnitude of the resultant displacement with the Pythagorean Theorem. Use tan to get the direction. 5. Write a proper statement sentence.

1.1 Speed and velocity Note Look at diagram on blackboard. x-comp: (Let E be “+” and W be “-”) → → → → → → Δdx = Δd1x+ Δd2x +Δd3x +Δd4x +Δd5x = 0 + 5.0km +7.5km(sin45º) +11.0km(sin70º) +8.8km(sin30º)

1.1 Speed and velocity Note → Δdx = +5.0km + 5.303km +10.337km +4.4km = +25.040 km = 25.040 km [E]

1.1 Speed and velocity Note y-comp: (Let N be “+” and S be “-”) → → → → → → Δdy = Δd1y+ Δd2y +Δd3y +Δd4y +Δd5y = 4.0km + 0 -7.5km(cos45º) +11.0km(cos70º) +8.8km(cos30º)

1.1 Speed and velocity Note → Δdy = 4.0km -5.303km +3.7622km +7.621km = +10.080 km = 10.080 km [N]

1.1 Speed and velocity Note Draw resultant vector diagram of the x-comp and the y-comp. c2 = a2 +b2 Δd2 = Δdx2 + Δdy2 (SCALARS)

1.1 Speed and velocity Note Δd2 = Δdx2 + Δdy2 (SCALARS) Δd = (Δdx2 + Δdy2)0.5 = [(25.040km)2 +(10.080km)2]0.5 = (627.00km2+101.61km2)0.5 = (728.61km)0.5 = 27.0km

1.1 Speed and velocity Note tan θ = Δdx Δdy = 25.040 km 10.080 km = 2.4841

1.1 Speed and velocity Note tan θ = 2.4841 θ = tan-1(2.4841) = 68.1º Therefore the resultant displacement of the ATV is 27.0km [N 68.1º E].

Speed Note: v = d/t means nothing!!!! Which v? Δv, v1, v2, vav, or vinst? vav = d/Δt where vav = average speed Δd = distance Δt = time

Velocity → → Δt = time vav = Δd/Δt → where vav = average velocity → → vav = Δd/Δt → where vav = average velocity Δd = displacement Δt = time

Velocity If the velocity is constant then at every point the velocity equals the average velocity. → → → → vav = v1 = v2 = vinst → also Δv = 0

1.1 Speed and velocity Note Example # 1: (1-D) A car drives for half an hour and covers a displacement of 20 km [W]. Find the average velocity of the car in m/s.

1.1 Speed and velocity Note Soln → vav = ? Δd = 20 km [W] = 20 000 m [W] Δt = 0.5hX60minX 60s = 1800s 1h 1 min

1.1 Speed and velocity Note → → vav = Δd Δt = 20 000 m [W] 1800 s = 11 m/s [W] Therefore the car had an average velocity of 11m/s [W].

1.1 Speed and velocity Note Example #3: (2-D) A car travels at 50.0 km/h [E] for 10.0 minutes, 80.0 km/h [N] for 25.0 minutes, and finally 100.0 km/h [40º S from W] for 35.0 minutes. a) Calculate the average speed in m/s. b) Calculate the average velocity in km/h and in m/s.

1.1 Speed and velocity Note Soln a) Calculate the displacement for each velocity. → → v = Δd/Δt → → Therefore Δd = v  Δt since velocity is constant for this interval we can use the above equation.

1.1 Speed and velocity Note → → Δd1 = v1  Δt = (50.0 km/h [E])( 10.0/60h) = 8.3333km [E] or 8333.3m [E] → → Δd2 = v2  Δt = (80.0 km/h [N])(25.0/60h) = 33.333 km [N] or 33333 m [N]

1.1 Speed and velocity Note → → Δd3 = v3  Δt = (100.0 km/h [40º S from W]) X (35.0/60h) = 58.333 km [S 50º W] or 58333 m [S 50º W]

1.1 Speed and velocity Note Δd = Δd1 + Δd2 + Δd3 = 8333.3m +33333m +58333m = 99999 m Δt = Δt1 + Δt2 + Δt3 = 10.0min + 25.0min + 35.0min = 70.0min = 70.0min X (60 s/1min) = 4200 s

1.1 Speed and velocity Note vav = Δd/ Δt = (99999 m)/(4200 s) = 23.8 m/s Therefore the average speed is 23.8 m/s.

1.1 Speed and velocity Note b) Draw a diagram of the displacements X-component: (Let E be “+” and let W be “–”) → → → → Δdx = Δd1x + Δd2x + Δd3x = + 8333.3m – 58333m(sin50º) = + 8333.3 m – 44686 m = – 36353 m = 36353 m [W]

1.1 Speed and velocity Note Y-component: (Let N be “+” and let S be “–”) → → → → Δdy = Δd1y + Δd2y + Δd3y = +33333m –58333m(cos50º) = + 33333 m – 37495.7 m = – 4162.7 m = 4162.7 m [S]

1.1 Speed and velocity Note Draw resultant vector diagram of the x-comp and the y-comp. Δdr2 = Δdx2 + Δdy2 (SCALARS) Δdr = (Δdx2 + Δdy2)0.5 Δdr = [(36353m)2 + (4162.7m)2]0.5 Δdr = (1.3215 X 109 m2 +1.7328 X 107 m2)0.5 Δdr = (1.3388 X 109 m2)0.5 Δdr = 36590 m

1.1 Speed and velocity Note tan θ = Δdx Δdy = 363532 m 4163 m = 8.7896 θ = tan-1(8.7896) = 83.5º

1.1 Speed and velocity Note → → vav = Δdr/Δt = (36590m [S 83.5º W])/(4200s) = 8.7121 m/s [S 83.5º W] or 31.4 km/h [S 83.5º W] Therefore the average velocity is 8.71 m/s [S 83.5º W] or 31.4 km/h [S 83.5º W].

Extra Practice Questions Practice #1: What is the resultant displacement of a jogger that goes 3.0 km [W], 1.0 km [E], 2.3 km [18º E from S], 1.5 km [63º S of E], and finally 2.2 km [N]?

Extra Practice Questions Practice #2: A boat travels at 20.0 km/h [NE] for 23.0 minutes, 40.0 km/h [N] for 35.0 minutes, and finally 50.0 km/h [40º N of W] for 55.0 minutes. a) Calculate the average speed in m/s. b) Calculate the average velocity in km/h and in m/s.

Extra Practice Questions Practice #3: A swimmer travels at 4.0 km/h [13º S of W] for 18.0 minutes, 3.5 km/h [70º W from S] for 21.0 minutes, and finally 3.0 km/h [25º N of W] for 11.0 minutes. a) Calculate the average speed in m/s. b) Calculate the average velocity in km/h and in m/s.

Extra Practice Questions Answers 1. 1.5 km [S 24.7º W] 2a) 11.3 m/s b) 34.7 km/h [N 27.0º W] or 9.64 m/s [N 27.0º W] 3 a) 0.99 m/s b) 3.4 km/h [S 80.8º W] or 0.95 m/s [S 80.8º W]