Physics 111 NMU Chapter 2 PDF

Motion in One & Two Dimension PHY111 Chapter II Hamdi M. Abdelhamid Physics Department Faculty of Science New Mansoura University 12/10/2024 Introduction Motion Everything moves! Motion is one of the...

Motion in One & Two Dimension PHY111 Chapter II Hamdi M. Abdelhamid Physics Department Faculty of Science New Mansoura University 12/10/2024 Introduction Motion Everything moves! Motion is one of the main topics in Physics I In the spirit of taking things apart for study, then putting them back together, we will first consider only motion along a Mans straight line. Simplification: Consider a moving object as a particle, i.e. it moves like a particle—a “point object” New Mans 12/10/2024 22  Introduction Kinematics: part of dynamics that motion while ignoring the external agents that might have caused or modified the motion ▪ For now, will consider motion in one dimension ▪ Along Describes motion a straight while ignoringline the external agents that might have caused or modified the motion▪ Motion represents a continual change in an object’s position. For now, will consider motion in one dimension Types Along of Motion a straight line Motion represents a continual change in an object’s position. ❑Translational ✓ An example is a car traveling on a highway. ❑Rotational ✓ An example is the Earth’s spin on its axis. ❑Vibrational ✓ An example is the back-and-forth movement of a pendulum. 12/10/2024 33 Introduction Basic Quantities in Kinematics 12/10/2024 44 Position Motion can be defined as the change of position over time. How can we represent position along a straight line? Position definition: ▪ Defines a starting point: origin (x = 0), x relative to origin ▪ Direction: positive (right or up), negative (left or down) ▪ It depends on time: t = 0 (start clock), x(t=0) does not have to be zero. Position has units of [Length]: meters. x = + 2.5 m x=-3m 12/10/2024 55 Quantities in Motion Any motion involves three concepts Displacement Velocity Acceleration These concepts can be used to study objects in motion. 12/10/2024 66 Displacement Displacement: is the change in position during some time interval. ❑ Displacement: x = x f (t f ) − xi (ti ) ▪ f stands for final and i stands for initial. ❑ It is a vector quantity. ❑ It has both magnitude and direction: + or - sign ❑ It has units of [length]: meters. x1 (t1) = + 2.5 m x2 (t2) = - 2.0 m Δx = -2.0 m - 2.5 m = -4.5 m x1 (t1) = - 3.0 m x2 (t2) = + 1.0 m Δx = +1.0 m + 3.0 m = +4.0 m 12/10/2024 77 Distance and Position-time graph ❑Displacement in space ✓ From A to B: Δx = xB – xA = 52 m – 30 m = 22 m ✓ From A to C: Δx = xc – xA = 38 m – 30 m = 8 m ❖Distance is the length of a path followed by a particle. ❑Distance is the length of a path followed by a particle ✓ from A to B: d = |xB – xA| = |52 m – 30 m| = 22 m ✓ from A to C: d = |xB – xA|+ |xC – xB| = 22 m + |38 m – 52 m| = 36 m ✓Displacement is not Distance. 12/10/2024 88  Velocity Velocity is the rate of change of position. Velocity is a vector quantity. Velocity has both magnitude and direction. Velocity has a unit of [length/time]: displacement meter/second. We will be concerned with three quantities, distance defined as: Average velocity vavg = x = x f − xi t t Average speed savg = total distance t Instantaneous velocity v = lim x = dx t → 0 t dt 12/10/2024 displacement 99 Average Velocity ❑ Average velocity x x f − xi vavg = = t t is the slope of the line segment between end points on a graph. ❑ Dimensions: length/time (L/T) [m/s]. ❑ SI unit: m/s. ❑ It is a vector (i.e. is signed), and displacement direction sets its sign. 12/10/2024 10 10 Average Speed ❑ Average speed total distance savg = t ❑ Dimension: length/time, [m/s]. ❑ Scalar: No direction involved. ❑ Not necessarily close to Vavg: ◼ Savg = (6m + 6m)/(3s+3s) = 2 m/s ◼ Vavg = (0 m)/(3s+3s) = 0 m/s 12/10/2024 11 11 Graphical Interpretation of Velocity Velocity can be determined from a position-time graph Average velocity equals the slope of the line joining the initial and final positions. It is a vector quantity. An object moving with a constant velocity will have a graph that is a straight line. 12/10/2024 12 12 Instantaneous Velocity ❑ Instantaneous means “at some given instant”. The instantaneous velocity indicates what is happening at every point of time. ❑ Limiting process: ◼ Chords approach the tangent as Δt => 0 ◼ Slope measure rate of change of position ❑ Instantaneous velocity: x dx v = lim = t → 0  t dt ❑ It is a vector quantity. ❑ Dimension: length/time (L/T), [m/s]. ❑ It is the slope of the tangent line to x(t). ❑ Instantaneous velocity v(t) is a function of time. 12/10/2024 13 13 Uniform Velocity ❑ Uniform velocity is the special case of constant velocity ❑ In this case, instantaneous velocities are always the same, all the instantaneous velocities will also equal the average velocity x x f − xi ❑ Begin with vx = = then x f = xi + vx t t t Note: we are plotting x v velocity vs. time x(t) v(t) xf vx xi 0 t 0 t ti tf 12/10/2024 14 14 Average Acceleration ❑ Changing velocity (non-uniform) means an acceleration is present. ❑ Acceleration is the rate of change of velocity. ❑ Acceleration is a vector quantity. ❑ Acceleration has both magnitude and direction. ❑ Acceleration has a dimensions of length/time2: [m/s2]. ❑ Definition: v v f − vi ◼ Average acceleration aavg = = t t f − t i ◼ Instantaneous acceleration v dv d dx d 2 v a = lim = = = 2 t → 0 t dt dt dt dt 12/10/2024 15 15 Average Acceleration Note: we are plotting velocity vs. time ❑ Average acceleration v v f − vi aavg = = t t f − t i ❑ Velocity as a function of time v f (t ) = vi + aavg t ❑ It is tempting to call a negative acceleration a “deceleration,” but note: ◼ When the sign of the velocity and the acceleration are the same (either positive or negative), then the speed is increasing ◼ When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing ❑ Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph 12/10/2024 16 16 Instantaneous and Uniform Acceleration ❑ The limit of the average acceleration as the time interval goes to zero v dv d dx d 2 v a = lim = = = t → 0 t dt dt dt dt 2 ❑ When the instantaneous accelerations are always the same, the acceleration will be uniform. The instantaneous acceleration will be equal to the average acceleration ❑ Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph 12/10/2024 17 17 Relationship between Acceleration and Velocity ❑ Velocity and acceleration are in the same direction ❑ Acceleration is uniform (blue arrows maintain the same length) ❑ Velocity is increasing (red arrows are getting longer) v f (t ) = vi + at ❑ Positive velocity and positive acceleration 12/10/2024 18 18 Relationship between Acceleration and Velocity (2nd Stage) ❑ Uniform velocity (shown by red arrows maintaining the same size) ❑ Acceleration equals zero v f (t ) = vi + at 12/10/2024 19 19 Relationship between Acceleration and Velocity (3nd Stage) ❑ Acceleration and velocity are in opposite directions ❑ Acceleration is uniform (blue arrows maintain the same length) ❑ Velocity is decreasing (red arrows are getting shorter) v f (t ) = vi + at ❑ Velocity is positive and acceleration is negative 12/10/2024 20 20  Kinematic Variables: x, v, a ❑ Position is a function of time: x = x(t ) ❑ Velocity is the rate of change of position. ❑ Acceleration is the rate of change of velocity. x dx v dv v = lim = a = lim = t → 0  t dt t →0 t dt d d dt dt ❑ Position Velocity Acceleration ❑ Graphical relationship between x, v, and a This same plot can apply to an elevator that is initially stationary, then moves upward, and then stops. Plot v and a as a function of time. 12/10/2024 21 21 Special Case: Motion with Uniform Acceleration ❑ Acceleration is a constant ❑ Kinematic Equations (which we will derive in a moment) v = v0 + at 1 x = v t = (v0 + v)t 2 x = v0t + 12 at 2 v = v0 + 2ax 2 2 12/10/2024 22 22 Derivation of the Equation 1 ❑ Given initial conditions: ◼ a(t) = constant = a, v(t = 0) = v0, x(t = 0) = x0 ❑ Start with definition of average acceleration: v v − v0 v − v0 v − v0 aavg = = = = =a t t − t0 t −0 t ❑ We immediately get the first equation v = v0 + at ❑ Shows velocity as a function of acceleration and time ❑ Use when you don’t know and aren’t asked to find the displacement 12/10/2024 23 23 Derivation of the Equation 2 ❑ Given initial conditions: ◼ a(t) = constant = a, v(t = 0) = v0, x(t = 0) = x0 ❑ Start with definition of average velocity: x − x0 x vavg = = t t ❑ Since velocity changes at a constant rate, we have 1 x = vavgt = (v0 + v)t 2 ❑ Gives displacement as a function of velocity and time ❑ Use when you don’t know and aren’t asked for the acceleration 12/10/2024 24 24 Derivation of the Equation 3 ❑ Given initial conditions: ◼ a(t) = constant = a, v(t = 0) = v0, x(t = 0) = x0 ❑ Start with the two just-derived equations: 1 v = v0 + at x = vavgt = (v0 + v)t 2 1 1 1 2 ❑ We have x = (v0 + v)t = (v0 + v0 + at )t x = x − x0 = v0t + at 2 2 2 ❑ Gives displacement as a function of all three quantities: time, initial velocity and acceleration ❑ Use when you don’t know and aren’t asked to find the final velocity 12/10/2024 25 25 Average Acceleration ❑ Given initial conditions: ◼ a(t) = constant = a, v(t = 0) = v0, x(t = 0) = x0 ❑ Rearrange the definition of average acceleration v v −, vto0 find the time v − v0 aavg = = =a t= t t a ❑ Use it to eliminate t in the second equation: − 2 (v + v )(v − v, )rearrange 0 to get 2 1 1 v v x = (v + v)t = 0 0 0 = 2 2a 2a v 2 = v0 + 2ax = v0 + 2a( x − x0 ) 2 2 ❑ Gives velocity as a function of acceleration and displacement ❑ Use when you don’t know and aren’t asked for the time 12/10/2024 26 26 Problem-Solving Hints ❑ Read the problem ❑ Draw a diagram ◼ Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations ❑ Label all quantities, be sure all the units are consistent ◼ Convert if necessary v = v0 + at ❑ Choose the appropriate kinematic equation ❑ Solve for the unknowns x = v0t + 12 at 2 ◼ You may have to solve two equations for two unknowns v 2 = v0 + 2ax 2 ❑ Check your results 12/10/2024 27 27 Example ❑ An airplane has a lift-off speed of 30 m/s after a take-off run of 300 m, what minimum constant acceleration? v = v0 + at x = v0t + 12 at 2 v 2 = v0 + 2ax 2 ❑ What is the corresponding take-off time? v = v0 + at x = v0t + 12 at 2 v 2 = v0 + 2ax 2 12/10/2024 28 28 Free Fall Acceleration y ❑ Earth gravity provides a constant acceleration. Most important case of constant acceleration. ❑ Free-fall acceleration is independent of mass. ❑ Magnitude: |a| = g = 9.8 m/s2 ❑ Direction: always downward, so ag is negative if we define “up” as positive, a = -g = -9.8 m/s2 ❑ Try to pick origin so that xi = 0 12/10/2024 29 29  Free Fall Example ❑ A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The stone just misses the edge of the roof on the its way down. Determine ❑ (a) the time needed for the stone to reach its maximum height. ❑ (b) the maximum height. ❑ (c) the time needed for the stone to return to the height from which it was thrown and the velocity of the stone at that instant. ❑ (d) the time needed for the stone to reach the ground ❑ (e) the velocity and position of the stone at t = Jan. 28-Feb. 1, 2013 5.00s 12/10/2024 30 30  Motion in Two Dimension ❑ Displacement and position in 2-D Reminder of vectors and vector algebra Displacement and position in 2-D ❑ Average and instantaneous velocity in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration in 2-D ❑ Average Projectile motion and instantaneous acceleration in 2-D Uniform circular motion Projectile motion ❑ velocity* Relative 12/10/2024 31 31  Motion in Two Dimension ❑Kinematic variables in one dimension ◼ Position: x(t) m Velocity: Reminder◼ of vectors and vectorv(t) m/s algebra x ◼ Acceleration: Displacement and positiona(t) m/s2 in 2-D Average and instantaneous velocity in 2-D ❑ Kinematic Average variables in three and instantaneous dimensions in 2-D acceleration y Projectile Position: r (t ) = xiˆ + yˆj + zkˆ m ◼ motion  Velocity: Uniform circular motion ◼ v (t ) = v x iˆ + v y ˆj + v z kˆ m/s  Acceleration: Relative velocity* ◼ a (t ) = a x iˆ + a y ˆj + a z kˆ m/s2 j i x ❑ All are vectors: have direction and magnitudes k z 12/10/2024 32 32  Position and Displacement ❑ In one dimension x = x2 (t 2 ) − x1 (t1 ) Reminderx1 of (t1)vectors = - 3.0 and m, x2vector (t2) = algebra + 1.0 m Δx = and Displacement +1.0 position m + 3.0 min= 2-D+4.0 m    r = r2 − r1 Average and instantaneous velocity in 2-D ❑ In two dimensions Average and instantaneous acceleration in 2-D ◼ Position: the position of an object is Projectile motion  described by its position vector r (t ) -- Uniformalways circular motion points to particle from origin. Relative velocity*    ◼ Displacement: r = r − r 2 1  r = ( x2iˆ + y2 ˆj ) − ( x1iˆ + y1 ˆj ) = ( x2 − x1 )iˆ + ( y2 − y1 ) ˆj = xiˆ + yˆj 12/10/2024 33 33  Average & Instantaneous Velocity  ❑ Average velocity  r vavg  t Remindervof x ˆ andy ˆ vector vectors = i + j = vavg , xiˆ +algebra vavg , y ˆj t position t avg Displacement and in 2-D ❑ Instantaneous Average and instantaneous velocityvelocity in 2-D Average andinstantaneous acceleration  in 2-D  r dr v  lim vavg = lim Projectile motion = t →0 t →0 t dt Uniform circular  motion  dr dx ˆ dy ˆ v= Relative velocity* = i+ j = vxiˆ + v y ˆj dt dt dt ❑ v is tangent to the path in x-y graph; 12/10/2024 34 34  Motion of a Turtle Reminder of vectors and vector algebra Displacement and position in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration in 2-D Projectile motion A turtle Uniform starts motion circular at the origin and moves with the speed of v0=10velocity* Relative cm/s in the direction of 25° to the horizontal. (a) Find the coordinates of a turtle 10 seconds later. (b) How far did the turtle walk in 10 seconds? 12/10/2024 35 35  Motion of a Turtle Notice, you can solve the equations independently for the horizontal Reminder (x) and of vectors vertical and vector (y) algebra components Displacement andofposition motion inand2-Dthen combine Average andthem! instantaneous velocity in 2-D    acceleration in 2-D v0 = v x + v y Average and instantaneous Projectile motion ❑ X components: Uniform circular v0 x = vmotion x = v0 xt = 90.6 cm 0 cos 25 = 9.06 cm/s Relative velocity* ❑ Y components: v0 y = v0 sin 25 = 4.23 cm/s y = v0 y t = 42.3 cm ❑ Distance from the origin: d = x 2 + y 2 = 100.0 cm 12/10/2024 36 36  Average & Instantaneous Acceleration  ❑ Average acceleration  v aavg  t Reminderaof v x ˆ and vectors vy ˆj vector = i + = aavg , x iˆ +algebra aavg , y ˆj t t avg Displacement and position in 2-D Average and instantaneous ❑ Instantaneous velocity in 2-D acceleration Average and instantaneous acceleration  in 2-D   v dv  dv dvx ˆ dvy ˆ a  lim aavg = lim Projectile motion = a= = i+ j = a x iˆ + a y ˆj t →0 t →0 t dt dt dt dt Uniform circular motion Relative velocity* ❑ The magnitude of the velocity (the speed) can change ❑ The direction of the velocity can change, even though the magnitude is constant ❑ Both the magnitude and the direction can change 12/10/2024 37 37  Summary in two dimension  ❑ Position r (t ) = xiˆ + yˆj   r x ˆ y ˆ ❑ Average velocity vavg = = i+ j = vavg , xiˆ + vavg , y ˆj Reminder of vectors and vectortalgebra t t Displacement and position in 2-D dx dy ❑ Instantaneous velocity v x  v y  Average and instantaneous velocity   in dt2-D dt  Average and instantaneous r dr dx ˆ in v (t ) = lim acceleration = i + 2-D dy ˆ = j = vxiˆ + v y ˆj t →0 t dt dt dt Projectile motion Uniform circular motion dvx d 2 x dvy d 2 y ❑ Acceleration ax  = 2 ay  = 2 Relative velocity* dt dt dt dt    v dv dvx ˆ dvy ˆ a (t ) = lim = = i+ j = a x iˆ + a y ˆj t →0 t dt dt dt    ❑ r (t), v (t ), and a (t ) are not necessarily same direction. 12/10/2024 38 38  Summary in two dimension ❑ Motions in each dimension are independent components ❑ Constant acceleration equations   and  vector    2 v = v0 + at Reminder of vectors r − r = v0t + 2 at algebra 1 Displacement and position in 2-D ❑ Constant Average acceleration velocity and instantaneous equationsin hold 2-D in each dimension Average and = v0 x + axt acceleration viny 2-D vx instantaneous = v0 y + a y t Projectile motion x − x = v Uniform circular0 motion 0x t + 1 a 2 x t 2 y − y 0 = v0y t + 1 a 2 y t 2 vx = v0 x + 2a x ( x − x0 ) Relative velocity* 2 2 v y = v0 y + 2a y ( y − y0 ) 2 2 ◼ t = 0 beginning of the process;  ◼ a = a x iˆ + a y ˆj where ax and ay are constant;   ◼ Initial velocity v0 = v0 x iˆ + v0 y ˆj initial displacement r0 = x0iˆ ;+ y0 ˆj 12/10/2024 39 39  Hints for solving problems ❑ Define coordinate system. Make sketch showing axes, origin. ❑ List known quantities. Find v0x , v0y , ax , ay , etc. Show Reminder of vectors and vector algebra initial conditions Displacement on sketch. and position in 2-D ❑ Listand Average equations of motion instantaneous to seeinwhich velocity 2-D ones to use. ❑ Time Average andt isinstantaneous the same for xacceleration and y directions. in 2-D Projectile x0 =motion x(t = 0), y0 = y(t = 0), v0x = vx(t = 0), v0y = vy(t = 0). Uniform circular motion ❑ Have an axis point along the direction of a if it is Relative velocity* constant. vx = v0 x + axt v y = v0 y + a y t x − x0 = v0 xt + 12 axt 2 y − y0 = v0 y t + 12 a y t 2 vx = v0 x + 2a x ( x − x0 ) 2 2 v y = v0 y + 2a y ( y − y0 ) 2 2 12/10/2024 40 40  Projectile Motion An object may move in both the x and y directions simultaneously. Assumptions Reminder of Projectile of vectors and vector Motion: algebra Displacement The free-fallandacceleration position in 2-D is constant over the Average rangeand instantaneous velocity in 2-D of motion. Average and instantaneous acceleration in 2-D It ismotion Projectile directed downward. This Uniform is the circular same as assuming a flat Earth over the motion rangevelocity* Relative of the motion. It is reasonable as long as the range is small compared to the radius of the Earth. The effect of air friction is negligible. 12/10/2024 41 41  Projectile Motion ❑ 2-D problem and define a coordinate system: x- horizontal, y- vertical (up +) ❑ Try to pick x0 = 0, y0 = 0 at t = 0 Reminder of vectors and vector algebra ❑ Horizontal motion + Vertical motion Displacement and position in 2-D ❑ Horizontal: ax = 0 , constant velocity motion Average and instantaneous velocity in 2-D ❑ Vertical: ay = -g = -9.8 m/s2, v0y = 0 Average and instantaneous acceleration in 2-D ❑ Equations: Projectile motion Horizontal Vertical Uniform circular motion vx = vvelocity* Relative 0 x + axt v y = v0 y + a y t x − x0 = v0 xt + 12 axt 2 y − y0 = v0 y t + 12 a y t 2 vx = v0 x + 2a x ( x − x0 ) v y 2 = v0 y 2 + 2a y ( y − y0 ) 2 2 12/10/2024 42 42  Projectile Motion ❑ X and Y motions happen independently, so we can treat them separately Reminder of = v0 x and vector v xvectors v y =algebra v0 y − gt Displacement and position in 2-D x = x 0 + v0 x t y = y + v t − 1 gt 2 Average and instantaneous velocity 0 in 02-D y 2 Average Horizontal and instantaneous acceleration Vertical in 2-D Projectile motion ❑ Try to pick x0 = 0, y0 = 0 at t = 0 Uniform circular motion ❑ Horizontal motion + Vertical motion Relative velocity* ❑ Horizontal: ax = 0 , constant velocity motion ❑ Vertical: ay = -g = -9.8 m/s2, ❑ x and y are connected by time t ❑ y(x) is a parabola 12/10/2024 43 43  Motion in Two Dimension ❑ 2-D problem and define a coordinate system. ❑ Horizontal: ax = 0 and vertical: ay = -g. ❑ Try to of Reminder pick x0 = 0, yand vectors 0 = 0 at t = 0. algebra vector ❑ Velocity initial Displacement andconditions: position in 2-D ◼ v0 can have x, y components. Average and instantaneous velocity in 2-D ◼ v0x is constant usually. v0 x = v0 cos  0 Average and instantaneous acceleration in 2-D ◼ v changes continuously. v0 x = v0 sin  0 Projectile0ymotion ❑ Equations: Uniform circular motion Horizontal Vertical Relative velocity* v x = v0 x v y = v0 y − gt x = x 0 + v0 x t y = y0 + v0 y t − 12 gt 2 12/10/2024 44 44  Trajectory of Projectile Motion Initial conditions (t = 0): x0 = 0, y0 = 0 ❑ v0x = v0 cosθ0 and v0y = v0 sinθ0 ❑ Horizontal Reminder of vectors motion: and vector algebra x x = 0 + Displacement andv0x t  t = position inv0 x2-D Average Vertical ❑ and motion: instantaneous velocity in 2-D Average andy = 0instantaneous + v0 y t − 12 gt 2 acceleration in 2-D 2 Projectileymotion  x  g x  = v0 y   −   Uniform circular  v0motion x  2  v0 x  Relative velocity* g y = x tan  0 − 2 x 2v0 cos  0 2 2 ❑ Parabola; ◼ θ0 = 0 and θ0 = 90 ? 12/10/2024 45 45  Range and Maximum Height of a Projectile ❑ Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0y = v0 sinθ0, then Reminder xof= vectors 0 + v0 xt and 0 =vector 0 + v0 y t −algebra 1 2 gt 2 h Displacement and2position v0 y 2v0 sinin  0 2-D t= = Average and instantaneous g g velocity in 2-D Average and instantaneous 2v0 cos  0v0 acceleration sin  0 v0 sin 2in 2-D 2 R = x − x0 = v0 x t = = 0 Projectile motion g g 2 Uniformh =circular motion t g  t  y − y0 = v0 y t h − 12 gth = v0 y −   2 Relative velocity* 2 2 2 Horizontal Vertical v0 sin 2  0 2 h= v y = v0 y − gt 2g v x = v0 x 2v0 y v y = v0 y − gt = v0 y − g = −v0 y x = x 0 + v0 x t y = y0 + v0 y t − 12 gt 2 g This equation is valid only for symmetric trajectory. 12/10/2024 46 46  Projectile Motion at Various Initial Angles Complementary values of the initial angle result in the v0 sin 2 2 sameofrange Reminder vectors and vector algebra R= Displacement and position in 2-D g Average and instantaneous velocity in 2-D The heights will be different Average and instantaneous acceleration in 2-D Projectile motion Thecircular Uniform maximum range occurs motion at avelocity* Relative projection angle of 45o 12/10/2024 47 47  Summary  ❑ Position r (t ) = xiˆ + yˆj   r x ˆ y ˆ ❑ Average velocity vavg = = i+ j = vavg , xiˆ + vavg , y ˆj Reminder of vectors and vectortalgebra t t Displacement and position in 2-D dx dy ❑ Instantaneous velocity v x  v  Average and instantaneous velocity indt2-D y dt    Average and instantaneous  v (t ) = lim acceleration = iˆ +in 2-D r dr dx dy ˆ = j = vxiˆ + v y ˆj t →0 t dt dt dt Projectile motion Uniform circular motion dvx d 2 x dvy d 2 y ❑ Acceleration ax  = 2 ay  = 2 Relative velocity* dt dt dt dt    v dv dvx ˆ dvy ˆ a (t ) = lim = = i+ j = a x iˆ + a y ˆj t →0 t dt dt dt    ❑ r (t), v (t ), and a (t ) are not necessarily in the same direction. 12/10/2024 48 48  Summary ❑ If a particle moves with constant acceleration a, motion equations are    2 rf = ri + vi t + 12 at Reminder of vectors and vector algebra r = x ˆ + y f ˆj = ( xi + v xit + 1 a xit 2 )iˆ + ( yi + v yi t + 1 a yi t 2 ) ˆj i Displacement and position in 2-D 2 f f 2 Average and instantaneousvvelocity   in 2-D = vi + at Average and instantaneous acceleration in 2-D  Projectile motionf (t ) = v fxiˆ + v fy ˆj = (vix + a xt )iˆ + (viy + a y t ) ˆj v Uniform circular motion Relative velocity* ❑ Projectile motion is one type of 2-D motion under constant acceleration, where ax = 0, ay = -g. 12/10/2024 49 49