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Vector Examples

Vector Examples. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB. A VECTOR describes anything that has both a MAGNITUDE and a DIRECTION The MAGNITUDE describes the size of the vector. The DIRECTION tells you where the vector is pointing.

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Vector Examples

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  1. Vector Examples Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  2. A VECTOR describes anything that has both a MAGNITUDE and a DIRECTION The MAGNITUDE describes the size of the vector. The DIRECTION tells you where the vector is pointing. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  3. V= 600 km/hr Here’s a typical example: An airplane is flying East at a velocity of 600 kilometers per hour. Here the magnitude (speed) is 600 km/hr and the direction is East. A diagram of this vector might look like this: Normally vectors will be written in BOLDFACE or with an arrow above the letter: = 600 km/hr East or V=600 km/hr East Notice that the diagram above has a non-bold V – this means magnitude. So V=600 km/hr describes the magnitude of V. You might also see this as |V| = 600 km/hr. The absolute value bars indicate magnitude. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  4. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  5. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  6. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr 100 km/hr Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  7. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr 100 km/hr So how do we find the new SPEED of the plane? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  8. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr 100 km/hr So how do we find the new SPEED of the plane? Answer: Add the vectors together Is the new speed just 600 km/hr +100 km/hr = 700 km/hr? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  9. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr 100 km/hr So how do we find the new SPEED of the plane? Answer: Add the vectors together Is the new speed just 600 km/hr +100 km/hr = 700 km/hr? NO Note: if the wind blows East, then we can simply add the numbers together. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  10. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr 100 km/hr Vtotal So how do we find the new SPEED of the plane? The new speed will be the magnitude of the SUM of the 2 vectors. In this case, the sum will look like the hypotenuse of a right triangle, labeled Vtotalabove. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  11. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr Vtotal 100 km/hr We need to use the Pythagorean Theorem to find the magnitude of Vtotal Vtotal≈ 608 km/hr Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  12. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr Vtotal 100 km/hr Vtotal≈ 608 km/hr Now we have the magnitude, but what is the direction? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  13. Let’s try a quick example with our airplane. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 600 km/hr Vtotal 100 km/hr Vtotal≈ 608 km/hr θ Now we have the magnitude, but what is the direction? We use our right-triangle rules for this: tan(θ) = opposite/adjacent = 100/600 θ = tan-1(1/6) θ ≈ 9.5° Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  14. Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane? 600 km/hr Vtotal 100 km/hr θ Vtotal≈ 608 km/hr θ ≈ 9.5° Finally! We have our answer! The plane will speed up to 608 km/hr, but be pushed off-course by an angle of 9.5°. Note: the angle may be described as 9.5° North of East. Next we will work out a (slightly harder) problem from a textbook. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  15. A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  16. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. The first step is to set up a coordinate system. The most convenient thing to do is just call East the x-direction and North the y-direction, as shown. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  17. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Now we look at each leg of the journey and find the components: First leg: This one is easy. 2.00 km in the x-direction; 0.00 km in the y-direction We could label everything to keep it all organized: If the first leg is represented by the vector V1, we can write the components separately. V1x= +2.00 km (note: positive indicates East – negative would be West) V1y= 0.00 km Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  18. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. The 2nd leg has components in both x- and y-directions, so we can find the components from our triangle rules: V2x = V2y = Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  19. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. The 2nd leg has components in both x- and y-directions, so we can find the components from our triangle rules: V2x = +3.50*cos(45°) ≈ 2.47 km (positive for East) V2y = -3.50*sin(45°) ≈ -2.47 km (negative for South) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  20. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. The 3rd leg has unknown components, but we can write them as unknowns: V3x = ? V3y = ? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  21. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. The next step is the only one where you have to think about the problem. Since we know that the total journey from start to finish is 5.80 km East, we know the components: Vtotal,x = Vtotal,y = Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  22. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. The next step is the only one where you have to think about the problem. Since we know that the total journey from start to finish is 5.80 km East, we know the components: Vtotal,x = +5.80 km Vtotal,y = 0.00 km Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  23. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Finally we can just add everything up and solve for our unknowns. We will have 2 equations – one for each direction: V1y + V2y + V3y = Vtotal,y V1x + V2x + V3x = Vtotal,x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  24. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Finally we can just add everything up and solve for our unknowns. We will have 2 equations – one for each direction: V1y + V2y + V3y = Vtotal,y 0.00 – 2.47 + V3y = 0.00 V3y = +2.47 km (positive is North) V1x + V2x + V3x = Vtotal,x 2.00 + 2.47 + V3x = 5.80 V3x = +1.33 km (positive is East) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  25. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3rd leg. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  26. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3rd leg. Use the Pythagorean Theorem to find the magnitude: V3 = 2.81 km For extra fun, find the angle of the vector too! Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  27. y x A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3rd leg. Use the Pythagorean Theorem to find the magnitude: V3 = 2.81 km Use tan(θ) = Vy/Vx to find the direction tan(θ) = 2.47/1.33 Θ = tan-1(1.86) Θ = 61.7° North of East (this angle is measured above the x-axis) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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