Physics Vector Study Guides

Chapter C Vectors and Two-Dimensional Kinematics Blinn College - Physics 2325 - Terry Honan C.1 - Vector Algebra - I Polar Coordinates x2 + y2 is the distance from the origin to the point and θ is the angle measured counterclockwise from the positive x axis to the point. The definitions of the trig functions, for general angles, are given by ( x, y) are the Cartesian (or rectangular) coordinates of some point on a plane. r and θ are the polar coordinates; r = y x y sin θ = , cos θ = and tan θ = . r r x Using the above definitions it is a straightforward matter to find the formulas for converting between polar and rectangular coordinates. y 5 ( x, y) 4 3 2 r θ 1 -5 -4 -3 -2 -1 1 2 3 4 5 x -1 -2 -3 -4 -5 x = -2 and y = 4 ⟹ r = 4.47 and θ = 116.6° Interactive Figure r and θ ⟹ x and y x = r cos θ and y = r sin θ (C.1) x and y ⟹ r and θ y r= x2 + y2 and θ =  tan-1  x  when x > 0 y 180 ° + tan-1  x  when x < 0 (C.2) The subtlety in solving for the angle follows from the identity: tan θ = tan(180 ° + θ). The range of the tan-1 (also written arctan) function is -90 ° < θ < 90 °, which corresponds to the quadrants I and IV; this is equivalent to the condition that x > 0. The case of quadrants II and III, when x < 0, requires shifting the result of the inverse tangent by 180 °. 2 | Chapter C - Vectors and Two-Dimensional Kinematics u Quadrant IV Quadrant I Quadrant II Quadrant III Quadrant IV y Quadrant II x<0 y r u1 tan-1 u2 θ 180°+ tan-1 u2 -90° Quadrant I x>0 -1 tan u1 90° 180° 270° 180°+ tan-1 u1 360° θ Quadrant III x<0 u2 x x Quadrant IV x>0 Example C.1 - Polar Coordinates Consider the point (x, y) = (-2, 4) in the Cartesian plane. Find the polar coordinates r and θ of this point. Solution This is a straightforward application of the formulas above, where x = -2 and y = 4. r= x2 + y2 = θ = 180 ° + tan-1 20 = 4.47 y x = 116.6 ° Vector Basics A vector is a quantity with both a magnitude and a direction. A scalar has only a magnitude; it is just a real number. The magnitude of a vector is a non-negative (positive or zero) scalar. Velocity is a vector quantity and speed is its magnitude. Acceleration, force and momentum are also vectors. Time, temperature, mass and pressure are examples of scalars. We will write a vector variable by a symbol with an "⇀" over it. The magnitude of a vector is given by the symbol without the arrow or by applying the " " brackets to the vector. If A is some vector then A = A is its magnitude. We can represent vectors by arrows. Suppose someone walks from a starting point P1 to a stopping point P2 . A displacement vector s (or Δ r) may be viewed as an arrow with its tail at P1 and its tip at P2 .  P2  s P1 A position vector r is a displacement vector with its tail at the origin. For more general vectors we represent them by arrows pointing in the direction of the vectors and the length of the arrow is proportional to the magnitude of the vector. For instance, if one velocity vector has a magnitude of 60 mi / hr and another has 30 mi / hr then the arrow representing the first should have twice the length of the second. A vector has no fixed position. If a vector arrow is moved keeping its length and direction fixed then it still is the same vector. A A Chapter C - Vectors and Two-Dimensional Kinematics | 3 Component Definition - Position and Displacement Vectors A position vector is a way to label a position in the Cartesian plane; it has its tail at the origin and its head at the position it labels. We will use an angled bracket notation for vectors. The position vector that labels the point (x, y) will be written as r = 〈x, y〉. y ( x,y) y r = 〈x, y〉 x x We will define the magnitude and direction of a vector so that the polar coordinates, r and θ, are the magnitude and direction of the vector. In the Cartesian plane we will denote vector from (x1 , y1 ) to (x2 , y2 ) by: s = 〈x2 - x1 , y2 - y1 〉 This is called a displacement vector. A position vector is clearly a displacement vector with its tail at the origin. P2 = (x2 , y2 ) s = 〈x2 -x1 , y2 -y1 〉 P1 = (x1 , y1 ) Component Definition - General Vectors We will write a 2-dimensional vector A as a pair of real numbers Ax and Ay called components. (A 3D vector is a triple.) We will use the "angled-bracket" notation for vectors. A = 〈 Ax , Ay 〉 The component Ax has the interpretation as the amount the vector A is in the x-direction and Ay the y-direction. y Ay A=〈Ax ,Ay 〉 x Ax Magnitude and Direction Angle We define Ax and Ay as the components of the vector A. Ax is the part of A in the x direction and similarly Ay is the y part. The Cartesian coordinates x and y are the components of a position vector r. For a two dimensional vector we can represent the direction with an angle, measured as in the polar coordinates. To convert between the magnitude and direction angle and the components of a two dimensional vector we have analogous expressions to the ones for polar coordinates. A and θ ⟹ Ax and A y Ax = A cos θ and Ay = A sin θ Ax and A y ⟹ A and θ (C.3) 4 | Chapter C - Vectors and Two-Dimensional Kinematics Ay A= 2 Ax + Ay 2 and θ =  tan-1  A  when Ax > 0 x Ay 180 ° + tan-1  A  when Ax < 0 (C.4) x Vector Addition Suppose a displacement vector s1 corresponds to someone walking from P1 to P2 . Suppose that then the person walks from P2 to P3 ; call this displacement s2 . The net displacement is the vector from P1 to P3 ; this is what we will define as the sum of the two displacements s1 + s2 . To generalize this to any vectors, we will define the sum of general vectors A and B. Draw the vectors as shown, with the tail of B at the tip of A. The sum the vectors A + B is the vector drawn from the tail of A to the tip of B. B A A+B With our component definition vector addition takes the very simple form: A + B = 〈Ax , Ay 〉 + 〈Bx , By 〉 = 〈Ax + Bx , Ay + By 〉. (C.5) Commutative Property An algebraic operation is commutative when changing the order of the items doesn't affect the result. For vector addition this takes the form. A+B=B+A The commutative property of the addition of reals implies this for vectors. B A A+B A B Because of the commutative property there are two more ways of adding vectors we can consider. In addition to placing the tail of B at the tip of A, we can place the tail of A at the tip of B. Also there is the parallelogram rule: Draw the two vector together tail to tail and complete the parallelogram; the sum it the vector from the common tail of the vectors to the opposite corner. A A+B B Associative Property From the definition of vector addition it is clear it satisfies. A + B + C = A + B + C This property is called associativity. For an associative operation it is not necessary to use brackets, since the order of the operation is unimportant. B A B+C A+B C (A + B)+ C = A + ( B + C ) Identity The identity vector 0 is the vector that leaves any other vector A unchanged under addition. Chapter C - Vectors and Two-Dimensional Kinematics | 5 A+0=A It is clear that the zero vector has zeros as components. 0 = 〈0, 0〉 The magnitude of the identity vector is 0, so we can write 0 =  0 . Note that the direction of 0 is undefined; in fact, it is the only vector with an undefined direction. Additive Inverse. For any vector A there is an additive inverse vector -A with the property: A + -A = 0 Clearly, this has the value -A = 〈-Ax , -Ay 〉 and has the same magnitude and is in the opposite direction. A -A Vector Subtraction We define vector subtraction by adding the inverse. A - B = A + -B In terms of components we have A - B = 〈Ax , Ay 〉 - 〈Bx , By 〉 = 〈Ax - Bx , Ay - By 〉. The simplest way to view the vector A - B is as the vector that when added to B gives A; if the vectors are drawn tail to tail then it is the vector from the tip of B to the tip of A A A-B B Scalar Multiplication If A is a vector and c is a scalar then we can define their product c A as a vector. c A = 〈c Ax , c Ay 〉 (C.6) It is clear that its magnitude is given by c A = c A, where c is the absolute value of the scalar. The direction of c A is the same as A when c > 0 and opposite to A when c < 0. When c = 0 we get 0 A = 0. Note also that 1 A = A and (-1) A = -A. The scalar multiplication operation has the associative and distributive properties. 6 | Chapter C - Vectors and Two-Dimensional Kinematics Associative Property (c d) A = c d A Distributive Properties (c + d) A = c A + d A and c A + B = c A + c B Unit Vectors and Notation A unit vector is a vector of magnitude one. We denote unit vectors with a " ^ " over its top. For any vector A we can simply find the unit  vector in its direction A by A  A= . A   Basis unit vectors are unit vectors along the coordinate axes. We will use x and y for the unit vectors in the x and y directions. The more   traditional notation for these unit vectors is to write them as i and j.     x = i = 〈1, 0〉 and y = j = 〈0, 1〉 Any vector can then be written in terms of these basis unit vectors   A = 〈 Ax , Ay 〉 = Ax x + Ay y y Ay  Ay y A=〈Ax ,Ay 〉  y  Ax Ax x  x x 3D Vectors   It is straightforward to generalize our two dimensional discussion to a three dimensional one. We need to add z = k for the unit vector in the z direction.       x = i = 〈1, 0, 0〉 , y = j = 〈0, 1, 0〉 and z = k = 〈0, 0, 1〉 We can similarly write any vector in terms or its components and unit vectors    A = 〈 A x , A y , A z 〉 = A x x + A y y + A z z. The magnitude of a vector is A = A = A x 2 + A y 2 + Az 2 . The only subtlety with three dimensions is specifying directions as angles. It takes two angles to specify a direction is 3D; these are the θ and ϕ of spherical coordinates. To avoid these issues the best way to represent a direction for a three dimensional vector is to just write a unit vector in the direction of that vector A  A= . A Example C.2 - Vector Addition A plane flies 500 km to the south and then 900 km in the direction 50° north of west. What is the plane’s net displacement? What are the magnitude and direction angle of the plane’s net displacement? Chapter C - Vectors and Two-Dimensional Kinematics | 7 North y 200km R x -600km East 900km B 500km A 50° -500km Interactive Figure Solution Here we are adding two vectors, starting with magnitude and direction information. To find the components using equation (C.3) we must identify the direction angles which are measured counterclockwise from the positive x direction, which we take to be East. Vector A is to the South so it has direction angle -90 ° and it has magnitude 500 km. Although we could use (C.3) find Ax and Ay it is easier to see that since it is purely in the negative y direction: A = 〈Ax , Ay 〉 = 〈0, -500 km〉 . The direction angle for vector B is θB = 180 ° - 50 ° = 130 °. (See the interactive diagram above.) The magnitude of B is B = 900 km so using (C.3) we get B = 〈Bx , By 〉 = 〈B cos θB , B sin θB 〉 = 〈-578.51, 689.44〉 km . The net displacement is sum of these two vectors, the resultant vector R = A + B. R = A + B = 〈R x , R y 〉 = 〈0, -500〉 km + 〈-578.51, 689.44〉 km = 〈-578.51, 189.44〉 km Using (C.4) we can find the Magnitude and direction angle of the net displacement. R= R2x + R2y = 609 km θ = 180 ° + tan-1 Ry Rx = 161.9 ° (Magnitude) (Direction Angle) The net displacement is just R. R = 〈-579, 189〉 km (Net Displacement) C.2 - Kinematics in 2D and 3D The General Problem Position as a Function of Time There are two equivalent ways of describing position in two dimensions. One is by labeling the coordinates (x, y). The other is by giving the position vector r of the position. The two are related; the coordinates are the components of the position vector. To label position as a function of time we can consider x and y as separate functions of time or as r as a function of time. 8 | Chapter C - Vectors and Two-Dimensional Kinematics  x(t) and y(t) ⟺ r(t) = 〈 x(t), y(t)〉 The actual path followed by the body is called the trajectory. It is represented by a plot of a path in the xy plane. y t r x x y t t Interactive Figure - Two dimensional motion may be visualized as vectors or as two separate cases of one dimensional motion, each representing a component of the vectors. Average Velocity The average velocity, as we saw in the one dimensional case, refers to two times. At ti the position vector is ri = r(ti ) and at t f it is r f = rt f . The displacement is the difference of these two positions Δ r = r f - ri = 〈x f - xi , y f - yi 〉. The average velocity vector is then defined as Δ r Δx Δy vave = = ,  = 〈vave,x , vave,y 〉. Δt Δt Δt (C.7) Chapter C - Vectors and Two-Dimensional Kinematics | 9 y ti vave  Δr tf ri r f x y x xi ti tf xf t yi ti vave,x = Δx Δt = slope tf vave,y = Δy Δt t = slope Interactive Figure - The average velocity can be found from Δ  r using  v ave = Δ  r / Δt. Instantaneous Velocity The instantaneous velocity is defined as the limit of the average velocity as Δ t approaches zero; it is the time derivative of the position vector. ⅆ r Δ r ⅆx ⅆy v = = lim = ,  = 〈vx , vy 〉. ⅆ t Δ t→0 Δ t ⅆt ⅆt (C.8) The magnitude of the velocity is called the speed. v =  v  = v2x + v2y = speed (C.9) 10 | Chapter C - Vectors and Two-Dimensional Kinematics y v vave  Δr x x y t t0 t t0 Interactive Figure - The instantaneous velocity is the limit of the average velocity as Δt → 0. Average Acceleration As with the one dimensional case acceleration is to velocity as the velocity is to position. aave = Δ v Δt = Δ vx Δ vy ,  = 〈aave,x , aave,y 〉. Δt Δt (C.10) y vf tf vi aave ti vi Δv x vf , you must redraw the velocities with a Interactive Figure - To find Δv  / Δt. common tail. The average acceleration can then be found  a ave = Δv Instantaneous Acceleration The instantaneous acceleration is similarly defined as a limit of the average acceleration or simply as the time derivative of the velocity. a= ⅆ v ⅆt = lim Δ t→0 Δ v Δt = ⅆ vx ⅆ vy ,  = 〈ax , ay 〉. ⅆt ⅆt (C.11) Chapter C - Vectors and Two-Dimensional Kinematics | 11 Interactive Figure - The instantaneous acceleration is the limit of the average acceleration as Δt → 0. Acceleration with Constant Speed The acceleration is the time derivative of the velocity. Constant velocity implies zero acceleration. Speed is the magnitude of the acceleration so there can be an acceleration if the speed is constant, because the direction can change. If the speed is constant then the acceleration must be perpendicular to the direction of motion, which is the direction of the velocity. constant speed ⟺ a ⊥ v If a car is driving at a constant speed and turning left then the acceleration is toward the left. Interactive Figure - For motion with constant speed, the acceleration is perpendicular to the velocity. Note that when the velocity vectors are redrawn with a common tail, their tips are on a circle with the speed as its radius. As Δt → 0, the average acceleration approaches the instantaneous acceleration which becomes perpendicular to the velocity. Parallel and Perpendicular Components of Acceleration If the speed is changing then there is a forward or backward component of the acceleration, where an increasing speed gives a forward component and a decreasing speed gives a backward component. If the speed and direction are changing then you have both components parallel and perpendicular to the direction of motion. Suppose a car is braking while turning left. There are then two components of the acceleration, the parallel component points backward and the perpendicular component points to the right. 12 | Chapter C - Vectors and Two-Dimensional Kinematics Constant Velocity and Acceleration The cases of constant velocity and acceleration follows the one dimensional example. Since we are now dealing with vectors the arbitrary constants introduced in antidifferentiation become vector quantities. const v ⟹ a = 0 and r(t) = r0 + v t 1 const a ⟹ v(t) = v0 + a t and r(t) = r0 + v0 t + a t2 2 For 2D motion with constant acceleration we can modify the four constant acceleration equations from Chapter B into two sets of four equations. x Equations vx = v0x + ax t Δx = 1 2 y Equations vy = v0y + ay t (vx + v0x ) t Δ x = v0x t + 1 2 ax t Δy = 2 1 2 (vy + v0y ) t Δ y = v0y t + v2x = v20x + 2 ax Δ x 1 2 ay t2 v2y = v20y + 2 ay Δ y. Projectile Motion An important case of motion with constant acceleration is that of projectile motion. Projectile motion is two dimensional motion of body under the influence of only gravity. If gravity is all that acts means that we ignore air resistance. Spin effects like a curving baseball or a slicing golf ball are associated with air resistance and will also be ignored. For projectile motion we will take x as the horizontal direction and y as the vertical direction. The acceleration due to gravity is downward and of magnitude g. Writing this as a vector gives  a = -g y = 〈0, -g〉 or ax = 0 and ay = -g Inserting these components into the two sets of equations above gives: Horizontal Vertical Equations Equations vx = v0x vy = v0y - g t Δ x = v0x t Δy = 1 2 (vy + v0y ) t Δ y = v0y t - 1 2 (C.12) g t2 v2y = v20y - 2 g Δ y. The horizontal motion is simple. Since there is no horizontal (the x direction) acceleration, the x component of the velocity is constant. The vertical part (the y direction) of the motion is equivalent to free fall. The key to solving projectile motion problems is keeping the two parts separate. Chapter C - Vectors and Two-Dimensional Kinematics | 13 y v0x v vy = 0 vy v v0x v0x vy v v v0y v0x θ x v0x vy v Interactive Figure - Projectile motion with a fixed initial speed and a varying launch angle. If a projectile is launched at an initial angle of θ with an initial speed v0 then the components of the initial velocity are given by v0x = v0 cos θ and v0y = v0 sin θ. Example C.3 - An Initial Horizontal Velocity A spring gun shoots a small ball with an initial horizontal velocity from a height of 1.35 m at the same time an identical ball is dropped from the same height. The ball that was shot lands a horizontal distance of 2.42 m from its initial position. (a) Which ball hits the ground first? Solution For both balls, the y-component of the initial velocity is zero. The shot ball’s velocity has an initial x-component and the dropped ball doesn’t. The time of fall is determined only by the vertical equations and the vertical motion of the two is identical. It follows that the two balls hit the floor at the same time. (b) What is the initial speed of the shot ball? Solution First define our variables in terms of the given information. h = 1.35 m , Δ x = 2.42 m and g = 9.80 m  s2 y h v0x = ? Δx The initial speed is also the x-component of the initial velocity. v0 = v20x + v20y = v20x + 02 = v0x x 14 | Chapter C - Vectors and Two-Dimensional Kinematics First we solve for its time of fall using the vertical equations using Δ y = -h. Δ y = v0y t - 1 2 g t 2 ⟹ -h = 0 - 1 2 g t2 ⟹ h = 1 2 2h g t2 ⟹ t = g = 0.52489 s From this we can use the horizontal equation to get v0x . Δ x = v0x t ⟹ v0x = Δ x / t = 4.67 m / s This is a very standard physics problem. In this class of projectile problems with an initial horizontal velocity, there are three variables: h, Δ x and v0x . You are two and asked for the third. (c) What was the speed of the ball when it hit the floor? Solution y h vx = v0x x Δx v=? vy The x-component of the balls velocity stays constant so vx = v0x . The speed is the magnitude of the velocity. v= v2x + v2y = v20x + v2y Using the time and the vertical equations we can find the y-component of the velocity. vy = v0y - g t = -g t = -5.1438 m / s and from this we find the speed. v= v20x + v2y = 6.95 m / s (d) What was the ball’s direction of motion when it hits the floor? Give the angle. Solution y h vx = v0x Δx x θ=? v vy In part (c) we found the magnitude of the velocity. Here we want the direction angle of the same velocity. Recall that the direction of motion is the direction of the velocity vector. θ = tan-1 vy v0x = -47.7 ° Chapter C - Vectors and Two-Dimensional Kinematics | 15 Example C.4 - Punted Football A punter kicks a football from ground level with an initial speed of 27 m / s at an initial angle of 63 °. (Neglect all air resistance.) x y R ymax R/2 T /2 T T /2 T t y ymax R/2 x R t (a) What is the maximum height reached by the football? Solution Begin with identifying what is given. We know the initial speed and initial angle. v0 = 27 m / s and θ = 63 ° We can calculate the components of the initial velocity. v0x = v0 cos θ = 12.258 m / s and v0y = v0 sin θ = 24.057 m / s The maximum height depends only on v0y . At the highest point the velocity is purely horizontal, so vy = 0. We are looking for ymax in the graphs above. ymax = Δ y when vy = 0. We need the fourth of the vertical projectile equations (C.12). v2y = v20y - 2 g Δ y ⟹ 0 = v20y - 2 g ymax ⟹ ymax = v20y 2g = 29.5 m (b) What is the “hang-time” of the punt? (Note that hang-time is the total time the football is in the air, from the ground to the ground.) Solution Since the football ends at the same elevation where it began, so Δ y = 0. We need to solve for time so the relevant formula is the third vertical projectile equation (C.12). Δ y = v0y t - 1 2 g t2 ⟹ 0 = v0y t - 1 2 g t2 = t v0y - 1 2 gt This gives two solutions but t = 0 is trivially true. We want the other. We will call this T as shown in the graphs above. 0 = v0y - 1 2 gT ⟹ T = 2 v0y g = 4.9096 s = 4.91 s (c) What is the horizontal range of the football? The range is the total horizontal distance the football travels in the air. Solution So far we have not used any horizontal information or equations. The range R is Δ x when Δ y = 0. Using the time T we just found in the second horizontal projectile equations (C.12). R = Δ x = v0x t = v0x T = 60.2 m To solve for the trajectory we can choose, for simplicity, that the motion begins at the origin x0 = 0 = y0 giving Δ x = x and Δ y = y. Solving the horizontal equation for time gives t = x / v0x . Inserting this into the vertical expression y = v0y t - y= It is clear that this is a parabola. v0y v0x x - g 2 v20x x2 . 1 2 g t2 gives 16 | Chapter C - Vectors and Two-Dimensional Kinematics The range R of a projectile is the total horizontal distance traveled in the air when it returns to its original level, y = 0 in the expression above. We can then factor out an x from the expression and solve for x. x = 0 and x = 2 v0x v0y 2 v0 cos θ v0 sin θ = g g = v20 g 2 cos θ sin θ. The x = 0 solution is trivial. Setting the other to R, writing the components in terms of v0 and θ, and using the identity sin 2 θ = 2 sin θ cos θ gives an expression for the range. v20 R= g (C.13) sin 2 θ Example C.5 - Punted Football (continued) (d) Apply the range formula (C.13) to the football example to get the same answer as in part (c). Solution R= v20 g sin 2 θ = 60.2 m C.3 - Relative Motion A frame of reference is some coordinate system used to study motion. Suppose there is a fixed frame K and a moving frame K ′ , that moves with a velocity v0 with respect to K. If we take a moving body to be at point P. If r is the vector from the origin of K to P, r′ is the vector from the origin of K ′ to P, and r0 is the vector from the origin of K to the origin of K ′ . P z′ r′ z r y′  r0 x′ y x Moving Frame K′ Fixed Frame K It follows that the three position vectors are related by r = r0 + r′ . We want to relate the velocities of the moving body with respect to these two frames. The velocities with respect to K and K ′ are v and v′ , respectively. We can relate the velocities in these two frames by taking the time derivative of the expression above: v = v0 + v′ , where v0 is the velocity of the moving frame. Example C.6 - Bubba Bubba drives his pickup at 85 mi / h while throwing his beer bottle out the window at 15 mi / h, relative to the pickup. (C.14) Chapter C - Vectors and Two-Dimensional Kinematics | 17 v′ 15mi/h v′ 15mi/h ? v v=? v0 85mi/h v v=? (a) θ v0 85mi/h (c) and (d) (a) What is the speed of the bottle, relative to the road, if the bottle is thrown forward? Solution Take the fixed frame K to be the frame of the road and K ′ to be the truck’s frame. We are studying the motion of the bottle with respect to these two frames. For a one dimensional problem remember that real numbers are one-dimensional vectors where the sign gives the direction. We then have (C.14) without the vector arrows. v = v0 + v′ = 85 mi / h + 15 mi / h = 100 mi / h (b) What is the speed of the bottle, relative to the road, if the bottle is thrown backward? Solution Now v′ is in the opposite direction, so we will make it negative. v = v0 + v′ = 85 mi / h + (-15 mi / h) = 70 mi / h (c) What is the speed of the bottle, relative to the road, if the bottle is thrown directly to Bubba’s left? Solution This is now two-dimensional. v0 and v′ are perpendicular and form two sides of a right triangle. v is the hypotenuse; to find its magnitude use the Pythagorean theorem. v= (85 mi / h)2 + (15 mi / h)2 = 86.3 mi / h (d) Relative to the road, what is the angle the bottle makes from the truck’s direction? Solution From the right triangle we my find θ using trig. θ = arctan 15 mi / h 85 mi / h = 10.0 ° Example C.7 - A Plane in a Cross-wind A plane flies to a city 800 mi to the east. In still air, the trip takes two hours. Suppose there is a strong 100 mi / h cross-wind blowing to the south. We will assume that the speed of the plane with respect to the air is fixed. (a) At what angle, north of east, must the plane aim to have a trajectory toward the east? Solution In relative motion problems we have v = v′ + v0 , where v is the velocity of some moving object relative to a fixed frame and v′ is the velocity of the object relative to a moving frame and v0 is the velocity of the moving frame. Here the moving object is the plane, the moving frame is the air and the ground is the fixed frame. The wind velocity is then v0 ; choosing east as the xdirection and north as the y-direction we have v0 = 〈0, -100〉 mi / h For v′ (the plane relative to the air) we can find its magnitude. 800 mi in two hours implies the speed of the plane relative to the air of v′ = 400 mi / h. As for the direction of v′ , it is directed at the unknown angle θ north of east. The velocity of the plane relative to the ground v is directed to the east with an unknown magnitude. 18 | Chapter C - Vectors and Two-Dimensional Kinematics For v′ (the plane relative to the air) we can find its magnitude. 800 mi in two hours implies the speed of the plane relative to the air of v′ = 400 mi / h. As for the direction of v′ , it is directed at the unknown angle θ north of east. The velocity of the plane relative to the ground v is directed to the east with an unknown magnitude. N ′ v W v′ = 4 /h 00 mi v0 v0 = 100 mi/h θ=? E v v=? S Given the right triangle above, we can solve for the angle θ. θ = sin-1 100 400 = 14.5 ° (b) How long does the trip take. Solution We know that the distance is d = 800 mi and d = v t, where v is the speed of the plane relative to the ground. To find that we use the Pythagorean theorem. v2 + v20 = v′ 2 ⟹ v = v′ 2 - v20 = (400 mi / h)2 - (100 mi / h)2 = 387.30 mi / h From this we can solve for the time. d=vt ⟹ t= d v = 2.07 h