Kinematics in Two Dimensions; Vectors Physics for Scientists & Engineers, 3rd Edition Douglas C. Giancoli Chapter 3 Kinematics in Two Dimensions; Vectors © Prentice Hall

Figure 3-1 Car traveling on a road. The green arrows represent the velocity vector at each position.

Figure 3-2 Combining vectors in one dimension

Figure 3-3 A person walks 10.0 km east and then 5.0 km north. These two disp0lacements are represented by the vectors D1 and D2, which are shown as arrows. The resultant displacement vector, DR, which is the vector sum of D1 and D2, is also shown. Measurement on the graph with ruler and protractor shows that DR has a magnitude of 11.2 km and points at an angle theta = 27 degrees north of east.

Figure 3-4 If the vectors are added in reverse order, the resultant is the same. (Compare Fig. 3-3)

Figure 3-5 (a) The three vectors in (a) can be added in any order with the same result, V1 + V2 + V3.

Figure 3-5 (b) In (b) VR = (V1 + V2) + V3, which clearly give the same result as in (c).

Figure 3-5 (c) VR = V1 + V2 + V3 without parentheses.

Figure 3-6 Vector addition by two different methods, (a) and (b). Part (c) is incorrect.

Figure 3-7 The negative of a vector is a vector having the same length but opposite direction.

Figure 3-8 Subtracting two vectors: V2 – V1.

Figure 3-9 Multiplying a vector V by a scalar c gives a vector whose magnitude is c times greater and in the same direction as V (or opposite direction if c is negative).

Figure 3-10 Resolving a vector V into its components along an arbitrarily chosen set of x and y axes. Note that the components, once found, themselves represent the vector. That is, the components contain as much information as the vector itself.

Figure 3-11 Finding the components of a vector using trigonometric functions, where θ is the angle with the x axis.

Figure 3-12 The components of V = V1 + V2 are Vx = V1x + V2x and Vy = V1y + V2y.

Figure 3-13 Example 3-1

Figure 3-14 Example 3-2

Figure 3-15 Unit vectors i, j, and k along the x, y, and z axes.

Figure 3-16 Path of a particle in the xy plane. At time t1 the particle is at point P1 given by the position vector r1; At time t2 the particle is at point P2 given by the position vector r2. The displacement vector for the time interval t2 – t1 is Δr = r2 – r1.

Figure 3-17 As we take Δt and Δr smaller and smaller [compare Fig. 3-16 and part (a) of this figure] we see that the direction of Δr and of the instantaneous velocity (Δr/ Δt, where Δt  0) is tangent to the curve at P1 (part b).

Figure 3-18 Velocity vectors v1 and v2 at instants t1 and t2 for the particle of Fig. 3-16. Direction of the average acceleration in this case, a avg = Δv / Δt.

Figure 3-20 Projectile motion. (A vertically falling object is shown at the left for comparison.)

Figure 3-22 Path of a projectile fired with initial velocity v0 at angle θ to the horizontal. Path is shown in black, the velocity vectors are green arrows, and velocity components are dashed.

Figure 3-23 Example 3-4

Figure 3-24 Example 3-5

Figure 3-25 Conceptual Example 3-6

Figure 3-26 Example 3-7

Figure 3-27 Example 3-8 The range R of a projectile; shows how generally there are two angles θ0, that will give the same range. Can you show that if one angle is θ01, the other angle is θ02 = 90° – θ01?

Figure 3-28 Example 3-9; The football leaves the punter’s foot at y = 0, and reaches the ground where y = -1.00 m.

Figure 3-29 Example 3-10

Figure 3-29 Example 3-10

Figure 3-31 A particle moving in a circle, showing how the velocity changes direction. Note that each point, the instantaneous velocity is in a direction tangent to the circular path.

Figure 3-32 (a) Determining the change in velocity, Δv, for a particle moving in a circle.

Figure 3-32 (b)

Figure 3-32 (c)

Figure 3-33 For uniform circular motion, a is always perpendicular to v.

Figure 3-34 The boat must head upstream at an angle θ if it is to move directly across the river. Velocity vectors are shown as green arrows: vBS = velocity of Boat with respect to the Shore, vBW = velocity of Boat with respect to the Water, vWS = velocity of Water with respect to the Shore (river current).

Figure 3-35 Derivation of relative velocity equation (Eq. 3-16), in this case for a person walking along the corridor in a train. We are looking down on the train and two reference frames are shown: xy on the Earth and x’y’ fixed on the train. We have rPT = position vector of person (P) relative to train (T) rPE = position vector of person (P) relative to Earth (E) rTE = position vector of train’s coordinate system (T) relative to Earth (E)

Figure 3-36 Example 3-13

Figure 3-37 Example 3-14: a boat heading directly across a river whose current moves at 1.20 m/s.

Figure 3-38 Example 3-15

Figure 3-39 Problem 8.

Figure 3-40 Problem 9.

Figure 3-41 Problems 12, 13, 14, and 15. Vector magnitudes are given in arbitrary units.

Figure 3-42 Problem 22.

Figure 3-43 Problem 29.

Figure 3-44 Problem 40.

Figure 3-45 Problem 41.

Figure 3-46 Problem 46.

Figure 3-47 Problem 47.

Figure 3-48 Problem 49. Given φ and v0, determine θ to make d maximum.

Figure 3-49 Problem 50.

Figure 3-50 Problem 59.

Figure 3-51 Problem 64.

Figure 3-52 Problem 66.

Figure 3-53 Problem 74.

Figure 3-54 Problem 75.

Figure 3-55 Problem 78.

Figure 3-56 Problem 83.

Figure 3-57 Problem 84.

Figure 3-58 Problem 87.

Figure 3-59 Problem 90.

Figure 3-60 Problem 93.

Figure 3-61 Problem 94.