Motion in Two Dimensions. 1.The Position, Velocity, and Acceleration Vectors 2.Two-Dimensional Motion with Constant Acceleration 3.

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1 Motion in Two Dimensions 1.The Position, Velocity, and Acceleration Vectors 2.Two-Dimensional Motion with Constant Acceleration 3.Projectile Motion

2 The position of an object is described by its position vector, r The displacement of the object is defined as the change in its position Δr = r f - r i

3 The average velocity is the ratio of the displacement to the time interval for the displacement v r t The direction of the average velocity is the direction of the displacement vector, Δr

4 The instantaneous velocity is the limit of the average velocity as Δt approaches zero The direction of the instantaneous velocity is along a line that is tangent to the path of the particle s direction of motion v lim t 0 r t dr dt

5 The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs. a vf vi v t t t f i

6 As a particle moves, Δv can be found in different ways The average acceleration is a vector quantity directed along Δv

7 The instantaneous acceleration is the limit of the average acceleration as Δv/Δt approaches zero a lim t 0 v t dv dt

8 Various changes in a particle s motion may produce an acceleration The magnitude of the velocity vector may change The direction of the velocity vector may change Even if the magnitude remains constant Both may change simultaneously

9 When the two-dimensional motion has a constant acceleration, a series of equations can be developed that describe the motion These equations will be similar to those of one-dimensional kinematics

10 Position vector Velocity Since acceleration is constant, we can also find an expression for the velocity as a function of time: v f = v i + at

11 The velocity vector can be represented by its components v f is generally not along the direction of either v i or at v f = v i + at

12 The position vector can also be expressed as a function of time: r f = r i + v i t + ½ at 2 This indicates that the position vector is the sum of three other vectors: The initial position vector The displacement resulting from v i t The displacement resulting from ½ at 2

13 Equation (4.8) and (4.9) in component form (because they are vector expressions) : t i f a v v t a v v t a v v y yi yf x xi xf (4.8a) i i f t t a v r r 2 y 2 1 yi i f 2 x 2 1 xi i f t a t v y y t a t v x x (4.9a)

14 Example (4.1) : Motion in a Plane A particle starts from the origin at t = 0 with an initial velocity having an x component of 20 m/s and a y component of 15 m/s. The particle moves in the xy plane with an x component of acceleration only, given by a x =4.0 m/s 2. (a) Determine the components of the velocity vector at any time and the total velocity vector at any time.

15 An object may move in both the x and y directions simultaneously The form of two-dimensional motion we will deal with is called projectile motion

16 Vertical Motion is the Same for Each Ball v ox v y v x v x v y 0 s 1 s 2 s v y v x v y 3 s v y v y

17 The free-fall acceleration g is constant over the range of motion And is directed downward The effect of air friction is negligible With these assumptions, an object in projectile motion will follow a parabolic path This path is called the trajectory

18 Consider the motion as the superposition of the motions in the x- and y-directions The x-direction has constant velocity a x = 0 The y-direction is free fall a y = -g The actual position at any time is given by: r f = r i + v i t + ½ gt 2

19 r f = r i + v i t + ½ g t 2 The final position is the vector sum of the initial position, the position resulting from the initial velocity and the position resulting from the acceleration

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21 The y-component of the velocity is zero at the maximum height of the trajectory The accleration stays the same throughout the trajectory

22 When analyzing projectile motion, two characteristics are of special interest The range, R, is the horizontal distance of the projectile The maximum height the projectile reaches is h

23 The maximum height of the projectile can be found in terms of the initial velocity vector: This equation is valid only for symmetric 2 2 motion v i sin h i 2g The range of a projectile can be expressed in terms of the initial velocity vector: This is valid only for symmetric trajectory R v 2 i sin 2 i g

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25 The maximum range occurs at i = 45 o Complementary angles will produce the same range The maximum height will be different for the two angles The times of the flight will be different for the two angles

26 Select a coordinate system Resolve the initial velocity into x and y components Analyze the horizontal motion using constant velocity techniques Analyze the vertical motion using constant acceleration techniques Remember that both directions share the same time

27 E 4.2 The Long Jump Angle 20, at the speed of 11m/s a) X =? b) Hmax =? E4.3 A Bull s-eye Every Time

28 Example (4.4) : That s Quite an Arm! A stone is thrown from the top of a building upward at an angle of 30.0 o to the horizontal and with an initial speed of 20.0 m/s, as shown in Figure (4.12). If the height of the building is 45.0 m, (a) how long is it before the stone hits the ground? (b) What is the speed of the stone just before it strikes the ground?

29 Example (4.5) : The End of the Ski Jump A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s, as shown in Figure (4.14). The landing incline below him falls off with a slope of 35.0 o. Where does he land on the incline?

30 25 m/s x y x = 50.0 m y = m

31 Example : A ball rolls off the top of a table 1.2 m high and lands on the floor at a horizontal distance of 2 m. What was the velocity as it left the table? What will be its speed when it strikes the floor? 1.2 m 2 m

32 v oy 28 m/s v y = 0 y max 30 o v ox v ox = 24.2 m/s v oy = + 14 m/s