Chapter 4 Motion in two dimensions (Chap. 2 in the textbook page 30) 1st week of March 2014/JuU 1435 Chapter 4 Motion in two dimensions (Chap. 2 in the textbook page 30) [Note à l'attention de l'instructeur Pour obtenir une aide détaillée sur la personnalisation de ce modèle, consultez la dernière diapositive. Consultez également le texte supplémentaire de la leçon dans le volet Commentaires de certaines diapositives.] Dr. Haykel Abdelhamid Elabidi 1

Units of Chapter 2 An introduction to vectors The velocity in two dimensions The acceleration in two dimensions 2

Scalars versus Vectors A scalar is a number with units. It can be positive, negative or zero. A vector is a mathematical quantity with both a magnitude and a direction Scalar quantities Masse Temperature Energy Power Speed Vector quantities Velocity Acceleration Displacement Force Field Your home School? 2 km To dscribe where the shool is located versus your home, It is not sufficient to give the distance between them. we have to give the distance and the direction. 2

The components of a vector A vector in xy plane can be resolved in two components.                       2

The components of a vector Example 2.1 page 30: A person walks 1 km due east. If the person then walks a second kilometer, what is the final distance from the starting point if the second kilometer is walked : (a) due east; (b) due west; (c) due south? We will call the first displacement A and the second B. Solution Example 2.1 page 30:  

The components of a vector Example 2.2 page 32: Find the components of the vectors A and B in Fig.2.7, if A = 2 and B = 3.

The components of a vector Solution Example 2.2 page 32: Ax and Ay Are positive: Ax = A cos  = 2 cos 30° = 2(0.866) = 1.73 Ay = A sin  = 2 sin 30° = 2 (0.500) = 1.00 From Fig.2.7b, Bx is positive and By is negative: cos 45° = sin 45° = 0.707, Bx = 3 cos 45° = 3 (0.707) = 2.12 By = -3 sin 45° = -3 (0.707) = -2.12

Adding and subtracting vectors Adding Vectors Using Components: Find the components of each vector to be added. Add the x- and y-components separately. Find the resultant vector.

Adding and subtracting vectors Subtracting Vectors: The negative of a vector is a vector of the same magnitude pointing in the opposite direction. Here

Adding and subtracting vectors Example 2.3 page 33: (a) Find the components of C = A + B. (b) Find the magnitude of C and its angle  with the x axis.

Adding and subtracting vectors Solution Example 2.3 page 33: Using the equation We can write Thus Cx = 6, and Cy = 8. (b) From the Pythagorean theorem: so C = 10. From Fig. 2.9, we see that the angle  satisfies.  = 53.1

Multiplying vector by scalar Multiplying vector by 3 increase its magnitude by a factor of 3, but does not change its direction.

The velocity in two dimensions In two dimensions, position, velocity, and acceleration are presented by vectors: motion in a plane. A problem involving motion in a plane is a pair of one-dimensional motion problems. If the displacement in a time interval ∆t is denoted by the vector ∆s, then the average velocity of the object is parallel to ∆s and is given by:

The velocity in two dimensions Example 2.4 page 34: A car travels halfway around an oval racetrack at a constant speed of 30 m s-1 (Fig. 2.11). (a) What are its instantaneous velocities at points 1 and 2? (b) It takes 40 s to go from 1 to 2, and these points are 300 m apart. What is the average velocity of the car during this time interval?

The velocity in two dimensions Solution 2.4 page 34: (a) The instantaneous velocity is tangent to the path of the car, and its magnitude is equal to the speed. Thus, at point 1 the velocity is directed in the +y direction and v1 = 30 ms-1 . Similarly, at point 2 the velocity is in the –y direction, and v2 = (30 m s-1)(- ) = -30 ms-1 b) The average velocity is the displacement divided by the elapsed time. The displacement is entirely along the x direction, so ∆s = 300m . Since ∆t = 40 s, The average velocity during this time interval is directed along the +x axis. Its magnitude is less than the speed of 30 m s-1 because the car does not travel in a straight line. The velocity in two dimensions

The velocity in two dimensions Example 2.5 page 33: A boat moves at 10 m s-1 relative to the water in a river. It is pointed toward the opposite shore. There is a 5 m s-1 current (Fig. 2.12a). Find the magnitude and direction of the boat’s velocity relative to the shore.

The velocity in two dimensions  

The acceleration in two dimensions Example 2.6 page 35: In Exercise 2.4 the velocity of the car changed from to in 40s. What was the average acceleration of the car in that time interval? The acceleration in two dimensions The average acceleration is:

The acceleration in two dimensions Solution Example 2.6 page 35: The average acceleration is defined as the velocity change divided by elapsed time: Thus the average acceleration during the time the car goes from point 1 to point 2 is directed in the –y direction, or downward in Fig. (2.11b)

The acceleration in two dimensions This exercise illustrates two important points: 1- If the velocity is constant, the acceleration is zero , since a is the rate of change of the velocity. However, when the speed is constant, the acceleration may or may not be zero. If an object moves at a constant speed along a curved path, its velocity is changing direction, and it is accelerating. We feel the effects of this acceleration when a car turns quickly. The acceleration is zero only when the speed and direction of motion are both constants. 2- The directions of the velocity and acceleration at any instant can be related in many ways. The magnitude and direction of a are determined by how v is changing. When a car moves along a straight road, the acceleration is parallel to the velocity if v is increasing and opposite if v is decreasing. When the motion is along a curved path, the acceleration is at some angle to the velocity.

Thank you for your attention Homeworks: Exercises 2.13.; 2.15.; 2.19 and 2.21. page 41 Thank you for your attention See you next time Inchallah 2