Chapter 2: Motion along a straight line 2.1: Position and displacement The location of an object is usually given in terms of a standard reference point, called the origin. The positive direction is taken to be the direction where the coordinates are increasing, and the negative direction as that where the coordinates are decreasing. Position of the body, x 1, x 2, x 3, A change in the coordinates of the position of the body describes the displacement of the body. For example, if the x-coordinate of a body changes from x 1 to x 2, then the displacement,  x = (x 2 -x 1 ). (Displacement is a vector quantity).

Average velocity (v avg ), is defined as the displacement (∆x) over the time duration time (∆t). or The SI unit of average velocity is (m/s) average velocity is vector quantity. 2.2: Average Velocity and Average Speed Average speed (s avg ), is defined as the ratio of the total distance traveled to the total duration time (∆t). The SI unit of average speed is (m/s) average speed is scalar quantity.

2.3: Average Velocity x 1 = – 4m, x 2 = 2m  x = x 2 – x 1  x = 2 – (– 4) = 6m t 1 = 1s, t 2 = 4s  t = t 2 – t 1  t = 4 – 1 = 3s Average velocity (v avg ) = (  x)/(  t) = (6/3) = 2 m/s The magnitude of the slope of the (x-t) graph gives the average velocity

Quick Quiz 2.1 (Page 27): Under which of the following conditions is the magnitude of the average velocity (v avg ) of a particle moving in one dimension smaller (  ) than the average speed (s avg ) over some time interval? (a) A particle moves in the +x direction without reversing. (b) A particle moves in the -x direction without reversing. (c) A particle moves in the +x direction and then reverses the direction of its motion. (d) There are no conditions for which this is true. (Answer C)

2.5: Instantaneous Velocity and Speed The instantaneous velocity (v, at  t  0) of a particle at a particular instant is the velocity of the particle at that instant (  t approaches a limiting value) The instantaneous velocity (v), is the slope of the tangent of the position-time graph at that particular instant of time. (Velocity (v) is a vector quantity). The instantaneous speed (magnitude of v) of a particle is defined as the magnitude of its instantaneous velocity. As with average speed, instantaneous speed has no direction (scalar quantity).

(A)Determine the displacement (x) of the particle in the time Intervals: t = 0 to t = 1 s and t = 1 s to t = 3 s. (B) Calculate the average velocity (v avg ) during these two time intervals. (C) Find the instantaneous velocity of the particle at t = 2.5 s. Example 2.3 (Page 30): Average and Instantaneous Velocity : x = – 4t + 2 t 2

(A) Determine the displacement (x) of the particle in the time Intervals: t = 0 to t = 1 s and t = 1 s to t = 3 s. at (t = 0 s): x = – 4 (0) + 2 (0) 2 = 0 m at (t = 1 s): x = – 4 (1) + 2 (1) 2 = – = – 2 m (negative x-direction) at (t = 3 s): x = – 4 (3) + 2 (3) 2 = – = 6 m (positive x-direction) Total displacement (  x) = 6 – (–2) = = 8 m x = – 4t + 2 t 2 (B) Calculate the average velocity (v avg ) during these two time intervals. average velocity (t =1s) = v avg (t = 1s) = (  x/  t) = (– 2/1) = – 2 m/s average velocity (t =3s) = v avg (t = 3s) = (  x/  t) = [6 – (– 2)]/(3 – 1) = (8/2) = 4 m/s (C) Find the instantaneous velocity (v) of the particle at (t = 2.5 s). x = – 4t + 2 t 2 v = (dx/dt) = – 4 + 4t v (t = 2.5 s) = – (2.5) = – = 6 m/s

2.6: Average (a avg ) and instant accelerations (a) accelerating (a) is the velocity of a particle changes with time. Average acceleration (a avg ): is the change of velocity over the change of time. As such, or The instantaneous (  t  0) acceleration (a, m/s 2 ) is defined as: or In terms of the position function, the acceleration can be defined as: The SI units for acceleration are m/s 2. Quick Quiz 2.2 (Page 32): If a car is traveling eastward and slowing down, what is the direction of the force on the car that causes it to slow down? (a)eastward (b) westward (c) neither of these. (Answer b)

Example 2.5 (Page 33): Average and Instantaneous Acceleration The velocity of a particle moving along the x-axis varies in time according to the expression v x = (40 – 5t 2 ) m/s, where (t) is in seconds. (A)Find the average acceleration in the time interval (t = 0 to t = 2.0 s). Solution v x = (40 – 5t 2 ) m/s v x1 (t = 0) = 40 – 5(0) 2 = 40 m/s v x2 (t = 2) = 40 – 5(2) 2 = 40 – 20 = 20 m/s  v x = v x2 – v x1 = 20 – 40 = – 20 m/s Average acceleration (a avg ) = (  v x /  t) = (– 20 / 2) = – 10 m/s 2

(B) Determine the instantaneous acceleration (a) at (t = 2.0 s) (Page 34) v x = (40 – 5t 2 ) m/s at (t = t +  t), then v x (t +  t) = 40 – 5 (t +  t) 2 v x (t +  t) = 40 – 5 (t 2 + 2t  t +  t 2 ) v x (t +  t) = 40 – 5 t 2 – 10 t  t – 5  t 2 v x (  t) = – 10 t  t – 5  t 2 a x (t = 2) = – 10 t = – 10 (2) = – 20 m/s 2

suppose a function (x) is proportional to some power of (t), such as: where (A) and (n) are constants, the derivation of (x) with respect to (t) is: Applying this rule to Example (2.5) – Page (33): v x = 40 – 5t 2 a x = (dv x /dx) = – 10t Quick Quiz 2.4 (Page 35): Which of the following is true? (a)If a car is traveling eastward, its acceleration is eastward. (b)If a car is slowing down, its acceleration must be negative. (c)A particle with constant acceleration can never stop and stay stopped. (Answer c) Evaluating the derivation of a function (Page 34):

Active Figure 2.9 (Page 35): velocity = constant acceleration = zero (Speeding up) velocity = increased acceleration = constant (v), (a) in the same direction (Slowing down) velocity = decreased acceleration = constant (v), (a) in the opposite direction

2.5 One-Dimensional Motion with Constant Acceleration (Page 36) V xf = v xi + a x t v xf 2 = v xi a x (x f – x i ) x f – x i = ½ (v xf + v xi ) t x f – x i = v xi t + ½ a x t 2 x f – x i = v xf – ½ a x t 2 Table 2.2: Kinematic Equations for Motion of a Particle Under Constant Acceleration (Page 38)

Quik Quiz 2.5 (Page 37) Answer (a)  (e) (b)  (d) (c)  (f) Quick Quiz 2.4 (Page 35): Which of the following is true? (a)If a car is traveling eastward, its acceleration is eastward. (b)If a car is slowing down, its acceleration must be negative. (c)A particle with constant acceleration can never stop and stay stopped. (Answer c)

Example 2.8 (Page 39): Watch Out for the Speed Limit! A car traveling at a constant speed of (v = 45.0 m/s) passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of (a = 3.00 m/s 2 ). How long (t = ?) does it take her to overtake the car? V(car) = 45 m/s A (trooper) = 3 m/s2 T (separation between car and trooper) = 1 s T (trooper to catch the car) = ? Distance of the car after (t = 1 s) x – x o = v o t + ½ a t 2 x – 0 = (45) (1) + ½ (0) (1) 2 = 45 m Trooper to catch the car X(trooper) = x(car) x o (trooper) + v o (trooper) t + ½ a(trooper) t 2 = x o (car) + v o (car) t + ½ a (car) t (0) t + ½ (3) t 2 = 45 + (45) t + ½ (0) t 2 (1.5) t 2 – (45) t – (45) = 0 t ≈ 31 s increasing the acceleration will decrease the time at which the trooper catches the car.

2.9: Free-Fall Acceleration In the absence of air resistance (Vacuum), all objects dropped toward the Earth with constant acceleration (g = ‒ 9.8 m/s 2 ). The value of (g) near the Earth’s surface decreases with increasing altitude. No external forces acting on the free falling objects, except for their weight. In vacuum, a feather and an apple will fall at the same rate.

Quick Quiz 2.6 (page 41): A ball is thrown upward. While the ball is in free fall, does its acceleration: (a)increase (b) decrease (c) increase and then decrease (d) decrease and then increase (e) remain constant? (Answer e) Quick Quiz 2.7 (page 41): After a ball is thrown upward and is in the air, its speed (a)increases (b) decreases (c) increases and then decreases (d) decreases and then increases (e) remains the same (Answer d)

Quick Quiz 2.8 (page 42): Which values represent the ball’s vertical velocity and acceleration at points (A), (C) and (E) in Figure 2.13a? (a) v y = 0, a y = – 9.80 m/s 2 (b) v y = 0, a y = 9.80 m/s 2 (c) v y = 0, a y = 0 (d) v y = – 9.80 m/s, a y = 0 (Answer a)