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## Chapter 1 Motion in a straight line

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**Chapter 1Motion in a straight line**1-2 Displacement vs Distance Average Velocity 1-3 INSTANTANEOUS VELOCITY 1 - 4 Acceleration 1.6 the acceleration of gravity and falling objects Norah Ali Almoneef**1-2 Displacement vs DistanceAverage Velocity**• Displacement is a vector that points from an object’s initial position to its final position • and has a magnitude that equals the shortest distance between the two positions. _Only depends on the initial and final positions • Independent of actual paths between the initial and final positions • Distance is a scalar • Depends on the initial and final positions as well as the actual path between them Norah Ali Almoneef**Displacement**This type of x(t) plot shows the position of an object at any time, e.g., x (m) Position at t=3 s, x(3) = 1 m 3 t (s) 4 Displacement between t=1 s and t=5 s Dx = 1.0 m - 2.0 m = -1.0 m -3 Norah Ali Almoneef**Given the train’s initial position and its final position**what is the displacement of the train? What is the distance traveled by the train ? Displacement = Norah Ali Almoneef**Example:**A boy travels from D to A,A to B .B to C.C to D Displacement from D to D ( which are initial and final points ) = 0 Distance traveled = 8 +4+8+4 = 24 m Norah Ali Almoneef**Example :**Distance = 4 m + 3 m =7 m Displacement = 5 m Norah Ali Almoneef**Speed and Velocity**The average speed being the distance traveled divided by the time required to cover the distance: How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m /s? Distance = 5400 s x 2.22 m / s = 11988 m Norah Ali Almoneef**Speed**Speed can be defined in a couple of ways: How fast something is moving The distance covered in a certain amount of time The rate of change of the position of an object Units for speed are: miles / hour (mi/hr) kilometers / hour (km/hr) feet / second (ft/s) This is the standard unitmeters / second (m/s) Norah Ali Almoneef**example**A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the velocity of the particle when t = 4 s. At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s Norah Ali Almoneef**example**What is Norah Ali Almoneef**example**From A to B What is B A Norah Ali Almoneef**1-3 INSTANTANEOUS VELOCITY**Instantaneous velocity – is how fast an object is moving at a particular instant. The position of a particle moving on the x axis is given by x = 7.8 + 9.2 t – 2.1t2. What is its instantaneous velocity at t = 3.5 seconds? v = 0 + 9.2 – (3)(2.1)t2 v = 0 + 9.2 – (3)(2.1)(3.5)= -68 m/s example Norah Ali Almoneef**1 - 4 Acceleration**Acceleration:is a rate at which a velocity is changing. Instantaneous acceleration = dv / dt = d2 x / d t2 Norah Ali Almoneef**Example**A car’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocity of 26 m/s. What is the acceleration of the car? The car is increasing its velocity by 8 m/s for every second it is moving. Norah Ali Almoneef**Instantaneous Acceleration**• Instantaneous Acceleration • Suppose a particle is moving in a straight line so that its position is given by the relationship x = (2.10 m/s2)t3 + 2.8 m. Find its instantaneous acceleration at 5 seconds. • v = dx / dt = (3)(2.1)t2 • a = dv / dt = (2)(3)(2.1)t at t= 5s • a = (2)(3)(2.1)(5) = 63 m/s2 Norah Ali Almoneef**example**Norah Ali Almoneef**example**A bullet train starts from rest from a station and travels along a straight horizontal track towards another station. The graph in fig. shows how the speed of the train varies withtime over the whole journey. Determine: (a) the total distance covered by the train, (b) the average speed of the train. A ) Total distance travelled Speed / ms-1 40 Time OR Total distance travelled = ‘area under the graph’ = (1/2)(10 + 16)(40) = 520 m 0 2 12 16 Average speed = (total distance) / (total time ) = 520 / 16 = 32.5 ms-1 Norah Ali Almoneef**example**: a car is traveling 30 m/s and approaches 10 m from an intersection when the driver sees a pedestrian and slams on his brakes and decelerates at a rate of 50 m/s2. (a) How long does it take the car to come to a stop? (b) how far does the car travel before coming to a stop? vf -vi= a t, where vo= 30 m/s, v = 0 m/s, and a = -50 m/s2 t = (0 -30)/(-50) = 0.6 s Δx= vit + ½ a t2= (30)(0.6) + ½(-50)(0.6)2= 18 -9 = 9 m Norah Ali Almoneef**1.5 finding the motion of an object**Equations of Kinematics for Constant Acceleration Norah Ali Almoneef**example**• How long does it take a car going 30 m/sec to stop of it decelerates at 7 m/sec2? Norah Ali Almoneef**example**- A car starting from rest attains a speed of 28 m/sec in 20 sec. Find the acceleration of the car and the distance it travels in this time. Norah Ali Almoneef**NO!!**If the object is slowing down, the acceleration vector is in the opposite direction of the velocity vector! 1. Velocity & accelerationare both vectors. Are the velocity and the acceleration always in the same direction? 2. Velocity & accelerationare vectors. Is it possible for an object to have a zero acceleration and a non-zero velocity? YES!! If the object is moving at a constant velocity, the acceleration vector is zero! Norah Ali Almoneef**Examples :**1 ) What is the acceleration of a car that increased its speed from 10 m/s to 30 m/s in 4 seconds? a = (30 m/s – 10 m/s) ÷ 4s = 20 m/s ÷ 4s = 5 m/s2 2)the same car now slows down back to 10 m/s in 5 seconds. What is his acceleration? a = (10 m/s – 30 m/s) ÷ 5s = (- 20 m/s) ÷ 5s = - 4 m/s2 Means slowing down Norah Ali Almoneef**Graphical Analysis**• *deduce from the shape of a speed-time graph when a body is: • (i) at rest • (ii) moving with uniform speed • (iii) moving with uniform acceleration • (iv) moving with non-uniform acceleration Velocity • (i) at rest • (ii) moving with uniform speed • (iii) moving with uniform acceleration • (iv) moving with non-uniform acceleration (ii) (iii) (iv) (i) Norah Ali Almoneef Time**example**Norah Ali Almoneef**A bus stopped at a bus-stop for 10 seconds before**accelerating to a velocity of 15 m/s in 4 seconds and then at a constant speed for the next 9 seconds. How does the graph look like? How far did the bus go in this 23 seconds? • Distance travelled in first 10 seconds is zero • Distance travelled in the next 4 seconds is • = ½ x 4 x 15 = 30 m • Distance travelled in the final 9 seconds is • = 9 x 15 = 135 m • Total distance travelled = 165 m Velocity /m/s (ii) 15 (iii) (i) 0 10 14 23 Time/s Norah Ali Almoneef**1.6 the acceleration of gravity and falling objects**Objects thrown straight up The acceleration of a falling object is due to the force of gravity between the object and the earth. Galileo showed that falling objects accelerate equally, neglecting air resistance. Galileo found that all things fall at the same rate. On the surface of the earth, in a vacuum, all objects accelerate towards the surface of the earth at 9.8 m/s2. The acceleration of gravity (g) for objects in free fall at the earth's surface is 9.8 m/s2. ( down ward ) g actually changes as we move to higher altitudes Norah Ali Almoneef**Equations of Kinematics for Constant Acceleration**For free fall Norah Ali Almoneef**A ball is dropped from a tall building and strikes the**ground 4 seconds later. A ) what velocity does it strike the ground B ) what distance does it fall? Norah Ali Almoneef**g = -**Norah Ali Almoneef**example**How high can a human throw a ball if he can throw it with initial velocity 90 m / h?. Norah Ali Almoneef**example**Norah Ali Almoneef**example**Norah Ali Almoneef**Notice in free fall**Norah Ali Almoneef**Word clues to numbers for problem solving**• “free-fall” acceleration due to gravity a=9.8m/s2, down • “at rest” not moving v=0 • “dropped” starts at rest and free-fall vi=0 and a=9.81m/s2, down • “constant velocity” no acceleration a=0 • “stops” final velocity is zero vf=0 Norah Ali Almoneef**Summary**2. Time interval: 1.Displacement: 2. Average velocity: 3.Instantaneous velocity: 4. Average acceleration: 5.nstantaneous acceleration: Norah Ali Almoneef