Mathematics
Problems involving Relative Velocity
Problems involving Relative Velocity deal with the motion of objects in relation to each other. These problems require the use of vector addition and subtraction to determine the velocity of one object relative to another. The key is to understand the direction and magnitude of the velocities involved.
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eBook - PDFEngineering Science N4 Student's Book
TVET FIRST
- MJJ van Rensburg(Author)
- 2022(Publication Date)
- Macmillan(Publisher)
1 Kinematics TVET FIRST 1 Module Kinematics Overview of Module 1 When you have completed this module, you should be able to: Unit 1.1: Relative velocity ● Solve problems dealing with constant linear motion analytically (Pythagoras or sine and cosine rules). ● Determine the relative velocity, shortest distance, time to intercept and actual velocity. Unit 1.2: Resulting velocity ● Calculate resulting velocity and direction of a maximum of two vectors. ● Calculate time to reach a certain destination. Unit 1.3: Projectiles ● Do calculations dealing with projectiles that are launched horizontally from a certain vertical height or launched at an angle from the horizontal landing on the same horizontal plane. ● Calculate maximum height reached by an object as well as time of flight and range. ● Calculate height and velocity at any part of the projectile path. ● Calculate velocity of projection. ● Calculate the angle of projection. Kinematics is the science that deals with motion irrespective of the force(s) that cause it. An example of kinematics would be to describe the motion of a motorcycle moving on a track in terms of its position, velocity, acceleration and time without considering the force from the engine or the effect of friction between the tyres and the track. In N1 to N3 you solved many problems involving motion. In N4 you will solve problems involving velocities of objects relative to each other. Starter activity Discuss the following in class: 1. What does kinematics mean? 2. Suppose you want to cross a river with a canoe and the water is flowing due south. In which direction should you steer the canoe if you want to travel at a 90° angle to the bank? 3. Car A is moving due east at 100 km/h. It is 2 km away from car B, which is on a road that crosses the first road. Car B is travelling due north at 60 km/h. Do you think it is possible to calculate the shortest distance between the two moving cars? 4.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
68 CHAPTER 3 Kinematics in Two Dimensions 3.4 Relative Velocity To someone hitchhiking along a highway, two cars speeding by in adjacent lanes seem like a blur. But if the cars have the same velocity, each driver sees the other remaining in place, one lane away. The hitchhiker observes a velocity of perhaps 30 m/s, but each driver observes the other’s velocity to be zero. Clearly, the velocity of an object is relative to the observer who is making the measurement. Figure 3.13 illustrates the concept of relative velocity by showing a passenger walking toward the front of a moving train. The people sitting on the train see the passenger walking with a velocity of +2.0 m/s, where the plus sign denotes a direction to the right. Suppose the train is moving with a velocity of +9.0 m/s relative to an observer standing on the ground. Then the ground-based observer would see the passenger moving with a velocity of +11 m/s, due in part to the walking motion and in part to the train’s motion. As an aid in describing relative velocity, let us define the following symbols: v → PT = velocity of the Passenger relative to the Train = +2.0 m/s v → TG = velocity of the Train relative to the Ground = +9.0 m/s v → PG = velocity of the Passenger relative to the Ground = +11 m/s In terms of these symbols, the situation in Figure 3.13 can be summarized as follows: v → PG = v → PT + v → TG (3.7) or v → PG = (2.0 m/s) + (9.0 m/s) = +11 m/s According to Equation 3.7*, v → PG is the vector sum of v → PT and v → TG , and this sum is shown in the drawing. Had the passenger been walking toward the rear of the train, rather than toward the front, the velocity relative to the ground-based observer would have been v → PG = (−2.0 m/s) + (9.0 m/s) = +7.0 m/s. Each velocity symbol in Equation 3.7 contains a two-letter subscript. The first letter in the subscript refers to the body that is moving, while the second letter indicates the object relative to which the velocity is measured.
eBook - PDF- Raymond Serway, Chris Vuille(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
Take careful note of the pattern of subscripts; rather than memorize Equation 3.11, it’s better to study the short derivation based on Figure 3.14. Note also that the equation doesn’t work for observers traveling a sizable fraction of the speed of light, when Einstein’s theory of special relativity comes into play. PROBLEM-SOLVING STRATEGY Relative Velocity 1. Label each object involved (usually three) with a letter that reminds you of what it is (e.g., E for Earth). 2. Look through the problem for phrases such as “The velocity of A relative to B” and write the velocities as v S AB . When a velocity is mentioned but it isn’t explic- itly stated as relative to something, it’s almost always relative to Earth. 3. Take the three velocities you’ve found and assemble them into an equation just like Equation 3.11, with subscripts in an analogous order. 4. There will be two unknown components. Solve for them with the x - and y - components of the equation developed in step 3. y x E B A r AB = r AE - r BE S S S r BE S r AE S Figure 3.14 The position of Car A relative to Car B can be found by vec- tor subtraction. The rate of change of the resultant vector with respect to time is the relative velocity equation. EXAMPLE 3.7 PITCHING PRACTICE ON THE TRAIN GOAL Solve a one-dimensional relative velocity problem. PROBLEM A train is traveling with a speed of 15 m/s relative to Earth. A passenger standing at the rear of the train pitches a baseball with a speed of 15 m/s relative to the train off the back end, in the direction opposite the motion of the train. (a) What is the velocity of the baseball relative to Earth? (b) What is the velocity of the baseball relative to the Earth if thrown in the opposite direction at the same speed? STRATEGY Solving these problems involves putting the proper subscripts on the velocities and arranging them as in Equation 3.11.
eBook - PDFTheoretical Mechanics for Sixth Forms
In Two Volumes
- C. Plumpton, W. A. Tomkys(Authors)
- 2016(Publication Date)
- Pergamon(Publisher)
CHAPTER V RELATIVE MOTION 5.1. Relative Position The position of a particle may be determined by reference to its displacement from an arbitrarily chosen point in space. The position thus determined is defined as the position of the particle relative to the chosen point. The frame of reference, by means of which the position of a particle is defined, is an essential feature of the definition. Thus, for example, the position of a particle in a moving railway carriage relative to another particle in the carriage may be constant, whereas the position of the first particle relative to a point on the ground outside the carriage changes with the motion of the carriage relative to the ground outside it. Because of the varied motions within the solar system, the position of a body in that system is considered relative to the sun. 5.2. Relative Velocity The velocity of a particle A relative to a particle B is defined as the rate of change of the displacement of A from B. Since the displacement of A from B is a vector, the velocity of A relative to B is also a vector. When particle A is moving with velocity u and particle B is moving with velocity v in the same direction, Fig. 5.1 (i), the component of the displacement of B from A in the direction of the motion is increasing at the rate (v— u). The component of displacement of B from A in a direc-tion at right angles to the direction of motion is constant. The velocity of B relative to A is therefore v—u in the direction of the motion. Similarly, when particle A is moving with velocity u and particle B is moving with velocity υ in the opposite direction, Fig. 5.1 (ii), the velocity of B relative to A is (v+u) in the direction of the motion of B. 106 RELATIVE MOTION 107 In each case the velocity of B relative to A is obtained by the vector addition of the velocity ofB to the reversed velocity of A.
- Paul Anthony Russell(Author)
- 2021(Publication Date)
- Reeds(Publisher)
If two objects, A and B, are moving on parallel courses at the same velocity as in Figure 2.13, the relative velocity of one to the other is zero. A typical example is two people sitting in the same railway carriage of a moving train; in the eyes of one, the other is not moving; the apparent or relative velocity of one to the other is zero. If, however, one object is moving exactly in the opposite direction to the other, such as two trains on outward and inward parallel tracks, each travelling at 50 km/h as illustrated in Figure 2.14, one appears to pass the other at 100 km/h; therefore, the relative velocity of one to the other is 100 km/h. A 50 km/h 50 km/h B ▲ Figure 2.13 Two objects moving in parallel in the same direction 54 • Applied Mechanics The relative velocities of objects moving on parallel courses are obvious and simple to understand, but when the courses are not parallel, it is necessary to draw vector diagrams. Consider a body A moving at 30 m/s due east and another body B moving at 35 m/s 20° north of east. A space diagram can be first sketched to show the absolute velocity of each; as these velocities are ‘relative to earth,’ they are marked A or B at the end behind the arrow and E (for earth) at the point end. See Figure 2.25. The vector diagram is now drawn with E as a common point for the two absolute velocities; the relative velocity of A to B, or B to A, is the vector connecting the two free ends. If the velocity of B relative to A is required, the arrow is put on pointing from B to A and shows how B appears to A as moving. If the velocity of A relative to B is required, the arrow is inserted in the direction A to B. This technique is known as vector subtraction. Example 2.16. Rain is falling at a velocity of 5 m/s. Find the velocity of the rain as it appears to a cyclist moving at 4 m/s.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
Sometimes, situations arise when two vehicles are in relative motion, and it is useful to know the relative velocity of one with respect to the other. Example 11 considers this type of relative motion. 3.4 | Relative Velocity 69 EXAMPLE 11 | Approaching an Intersection Figure 3.16a shows two cars approaching an intersection along perpendicular roads. The cars have the following velocities: v B AG 5 velocity of car A relative to the Ground 5 25.0 m/s, eastward v B BG 5 velocity of car B relative to the Ground 5 15.8 m/s, northward Find the magnitude and direction of v B AB , where v B AB 5 velocity of car A as measured by a passenger in car B Reasoning To find v B AB , we use an equation whose subscripts follow the order outlined earlier. Thus, v B AB 5 v B A G 1 v B G B In this equation, the term v B GB is the velocity of the ground relative to a passenger in car B, rather than v B BG , which is given as 15.8 m/s, northward. In other words, the subscripts are reversed. However, v B GB is related to v B BG according to v B GB 5 2 v B BG This relationship reflects the fact that a passenger in car B, moving northward relative to the ground, looks out the car window and sees the ground moving southward, in the opposite direc- tion. Therefore, the equation v B AB 5 v B AG 1 v B GB may be used to find v B AB , provided we recognize v B GB as a vector that points opposite to the given velocity v B BG . With this in mind, Figure 3.16b illustrates how v B AG and v B GB are added vectorially to give v B AB . Solution From the vector triangle in Figure 3.16b, the magnitude and direction of v B AB can be calculated as v AB 5 2(v AG ) 2 1 (v GB ) 2 5 2(25.0 m/s) 2 1 (215.8 m/s) 2 5 29.6 m/s and cos u 5 v AG v AB or u 5 cos 21 a v AG v AB b 5 cos 21 a 25.0 m/s 29.6 m/s b 5 32.4° Problem-Solving Insight In general, the velocity of object R relative to object S is always the negative of the velocity of object S relative to R: v B RS 5 2 v B SR .
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
But if the cars have the same velocity, each driver sees the other remaining in place, one lane away. The hitchhiker observes a velocity of perhaps 30 m/s, but each driver observes the other’s velocity to be zero. Clearly, the velocity of an object is relative to the observer who is making the measurement. Figure 3.13 illustrates the concept of relative velocity by showing a passenger walking toward the front of a moving train. The people sitting on the train see the passenger walking with a velocity of 12.0 m/s, where the plus sign denotes a direction to the right. Suppose the train is moving with a velocity of 19.0 m/s relative to an observer standing on the ground. Then the ground-based observer would see the passenger moving with a velocity of 111 m/s, due in part to the walking motion and in part to the train’s motion. As an aid in describing relative velocity, let us define the following symbols: v B PT 5 velocity of the Passenger relative to the Train 512.0 m/s v B TG 5 velocity of the Train relative to the Ground 5 19.0 m/s v B PG 5 velocity of the Passenger relative to the Ground 5 111 m/s In landing on a moving aircraft carrier, the pilot of the helicopter must match the helicopter’s horizontal velocity to the carrier’s velocity, so that the relative velocity of the helicopter and the carrier is zero. TAL COHEN/AFP/Getty Images v TG = +9.0 m/s Ground-based observer v PT = +2.0 m/s v PG = +11.0 m/s v TG v PT Figure 3.13 The velocity of the passenger relative to the ground-based observer is v B PG . It is the vector sum of the velocity v B PT of the passenger relative to the train and the velocity v B TG of the train relative to the ground: v B PG 5 v B PT 1 v B TG . 62 Chapter 3 | Kinematics in Two Dimensions Current Current (a) (b) Figure 3.14 (a) A boat with its engine turned off is carried along by the current. (b) With the engine turned on, the boat moves across the river in a diagonal fashion.
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