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Kinematics concepts

Kinematics concepts . 3UB Physics. Average Speed twist. A car travels 1.0 km through a town at 50.0 km/h and then drives 1.0 km through the country at 100.0 km/h. What is the average speed of this car? v = Δd / Δt = total distance/total time

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Kinematics concepts

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  1. Kinematics concepts 3UB Physics

  2. Average Speed twist • A car travels 1.0 km through a town at 50.0 km/h and then drives 1.0 km through the country at 100.0 km/h. What is the average speed of this car? • v = Δd/Δt • = total distance/total time • We need the time for each section of the motion!

  3. Average Speedtwist t1= d1/v1 = 1.0 km/50.0 km/h = 0.020h t2 = d2/v2 = 1.0 km/100.0km/h = 0.010h

  4. Average Speedtwist v = (1.0 + 1.0)km/(0.020 + 0.010)h = 67 km/h • Note: This average speed is NOT the average of 100 km/h and 50 km/h. Why?

  5. Graphing this question • The car drove 1.0 km at 50 km/h in 0.20 h and then 1.0 km at 100 km/h for 0.10 h. The average speed we found was 67 km/h. The two motions are each UM:

  6. UM question 1 • The following graph shows the Northward motion of a wolf tracking a biology teacher.

  7. UM question 1 • A) What is the wolf’s position at 4.0 s? • b) What is the wolf’s displacement between 2.0 s and 8.0 s? • c) What is the velocity of the wolf?

  8. UM question 1 • a) 4.0 m [N] • b) • Δd =df-di • = 8.0 m [N] – 2.0 m [N] • = 6.0 m [N] • c) • v = Δd/Δt • =(10m[N] – 0)/(10s – 0) • = 1.0m/s [N

  9. UM question 2 • The motion of a biology teacher heading North during migration season. The origin represents a town he passes through.

  10. UM question 2 • a) What is the total displacement? • b) What is the total distance traveled? • c) What is the total displacement from the town? • d) At what time does the migrating dude reach the town? • e) Where does he start? • f) What is his average velocity?

  11. UM question 2 • a) Δd = df-di • = 60 km – (-40 km) = 100 km [N] • b) Δd = 100 km • c) Δd from town = 60 km [N] • d) Δt = 4 h • e) He starts 40 km [S] of town. • f) v = Δd/Δt = (60 km – 0)/(10 h – 4 h) • = 10 km/h [N]

  12. UM Problem 3 • An aimless chemistry teacher wanders as:

  13. UM Problem 3 • Notice that we must label displacement rather than distance; why? • The slopes for each line segment, yields the velocity for that time interval. Find each. • Describe the motion of the above graph. • What section shows the highest velocity? • Which section shows the greatest displacement? Find Δd for each section. • What is the total displacement of the object?

  14. Displacement and distance from v-t graphs. • For motion comprised of UM and UAM, we will need to find the total distance and displacement by adding the areas up.

  15. Displacement and distance from v-t graphs. • The total distance is found by adding areas. • The total displacement is found by adding areas above the x-axis and adding negative areas below the x-axis. • Break up the graph into a series of triangles and rectangles to find the distance or displacement at certain times. (Useful info if asked to plot a d-t graph for the same motion!) • In this question, find: distance, displacement, speed and velocity. • Sketch the distance and displacement versus time graphs

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