Physics Answer Note #49

Question 1 Add the two following vectors in rectangular notation; A = 26i + 4j and B = 9i + –29j. Answer To add two vectors in rectangular notation, we simply add the corresponding components of the vectors. In this case, vector A has a horizontal component of 26i and a vertical component of 4j, while vector B has a horizontal component of 9i and a vertical component of -29j. When adding the horizontal components, 26i + 9i = 35i. Similarly, the vertical components will yield 4j + (-29j) = -25j. Therefore, the resultant vector, R = A + B, can be represented as R = 35i - 25j. We can also visualize this addition as moving from the initial point (0,0) to the endpoint of vector A, and then from that endpoint to the endpoint of vector B. The final position will be the endpoint of the resultant vector R. Graphical representation of vector addition: Source: http://electromagnetism.physicscentral.com/electro/vector/3.php Overall, the process of adding two vectors in rectangular notation is straightforward and can be easily visualized or solved mathematically. By understanding the basics of vector addition, we can apply this concept to various fields such as physics, engineering, and mathematics. Question 2 find the area under the velocity time graph (the area between the line and the x-axis). what does this measure Answer The area under the velocity time graph, also known as the velocity-time integral or the distance traveled, is a measure of the displacement of an object over a given time interval. It represents the total distance that an object has traveled in a specified time period. To understand this concept further, let's first define velocity and time. Velocity is the rate of change of position of an object, while time is the duration of an event. A velocity-time graph measures the velocity of an object at different points in time, as shown by the slope of the line. The area under this graph is the displacement of the object, which can be calculated by finding the integral of the velocity-time function. Question 3 Convert vector A = sa002-1.jpg in polar notation to rectangular notation in i & j notation. round your values to 1 decimal place. Answer Firstly, we need to understand the concept of polar and rectangular notations. Polar notation, also known as polar coordinates, represents a point in a plane by its distance from the origin (r) and the angle it makes with the positive x-axis (θ). On the other hand, rectangular notation, also known as the Cartesian coordinates, represents a point in a plane by its horizontal (x) and vertical (y) distance from the origin. To convert vector A from polar notation to rectangular notation, we will use the following formulas: x = r cos(θ) y = r sin(θ) Firstly, we need to find the distance from the origin (r) by using the Pythagorean theorem: r = √(x² + y²) Next, we need to find the angle (θ) by using the inverse tangent function: θ = tan-1 ( y / x ) Now, let's apply these formulas to convert vector A from polar to rectangular notation: Given: r = 5, θ = π/4 Step 1: Finding x and y x = 5 cos(π/4) x = 5 * 0.707 (using the value of cosine of 45 degrees from a trigonometric table or calculator) x = 3.5 (rounded to 1 decimal place) y = 5 sin(π/4) y = 5 * 0.707 (using the value of sine of 45 degrees from a trigonometric table or calculator) y = 3.5 (rounded to 1 decimal place) Step 2: Writing in rectangular notation in i & j form Therefore, the rectangular notation of vector A is (3.5i + 3.5j). Question 4 Find the horizontal component of vector A where A = 396angle48. Round your answer to the nearest whole number - no decimal Answer To find the horizontal component of a vector, we will use trigonometric functions such as sine and cosine. First, we need to understand that the angle given in the problem, 48 degrees, is measured from the positive x-axis in a counterclockwise direction. This means that the vector A is pointing in the Northeast direction. To find the horizontal component, we will use the cosine function which is defined as the ratio of the adjacent side (in this case, the horizontal component) to the hypotenuse (in this case, the magnitude of vector A). Therefore, we can write the following equation: cos(48) = Adjacent Side / A To find the horizontal component, we need to isolate the adjacent side by multiplying both sides of the equation by A. cos(48) * A = Adjacent Side Now, we can substitute the values given in the problem. The magnitude of vector A is given as 396, so the equation becomes: cos(48) * 396 = Adjacent Side Using a scientific calculator, we can find the value of cos(48) which is approximately 0.6691. Multiplying this value by 396 gives us the adjacent side which is approximately 265.0308. Since the question asks for the nearest whole number without any decimals, we will round the result to the nearest whole number. Therefore, the horizontal component of vector A is 265. Question 5 The airplane velocity "v" = 383 m/s is shown in the diagram above, pointing to the right. The actual direction the plane is going is shown by the red arrow, as it is being pushed off course by the wind, "w" shown by the green arrow, coming from the top above at 75 m/s at an angle, "a" = 66 degrees. What is the angle theta, between the heading (the direction the plane is pointing toward) and the actual direction (red arrow)? Answer In order to accurately determine the angle theta between the heading and the actual direction of the airplane, we must first understand the concepts of vector addition and trigonometry. Vector addition is the process of combining two or more vectors to determine the resultant vector. This can be represented graphically by placing the vectors' tails at the same point and drawing a line from the starting point of the first vector to the ending point of the last vector. The direction and magnitude of this line, or the resultant vector, will be the sum of the original vectors. In this scenario, we have two vectors - the velocity of the airplane, represented by "v" = 383 m/s, and the wind speed, represented by "w" = 75 m/s at an angle of "a" = 66 degrees. Using vector addition, we can determine the resultant vector, which will represent the actual direction of the airplane. To do this, we must first resolve the wind vector into its horizontal and vertical components. This can be done using trigonometric functions. The horizontal component, or x-component, of the wind vector will be given by multiplying the magnitude of the vector by the cosine of the angle, while the vertical component, or y-component, will be given by multiplying the magnitude of the vector by the sine of the angle. Hence, the horizontal component of the wind vector is 75 m/s * cos(66 degrees) = 30.82 m/s, while the vertical component is 75 m/s * sin (66 degrees) = 70.76 m/s. Now, we can add these components to the x-component of the velocity vector, as both vectors are acting in the same direction (to the right). This results in a resultant xcomponent of 383 m/s + 30.82 m/s = 413.82 m/s. Next, we can add the y-components of the velocity and wind vectors, taking into account their respective directions. The y-component of the velocity vector is 0, as it is not acting in the vertical direction, while the y-component of the wind vector is acting in the opposite direction, hence it will be subtracted. This gives us a resultant y-component of 0 - 70.76 m/s = -70.76 m/s. Using these x and y components, we can now determine the magnitude and direction of the resultant vector using the Pythagorean theorem and inverse tangent function, respectively. The magnitude of the resultant vector can be calculated as the square root of the sum of the squares of the x and y components. Hence, the magnitude is √(413.82 m/s)^2 + (70.76 m/s)^2 = 420.52 m/s. To determine the direction of the resultant vector, we can use the inverse tangent function, which is defined as the ratio of the opposite side to the adjacent side of a right triangle. In this case, the opposite side is the y-component (-70.76 m/s) and the adjacent side is the x-component (413.82 m/s). Hence, the inverse tangent of -70.76 m/s divided by 413.82 m/s is -9.7 degrees. However, this value is the direction of the resultant vector with respect to the horizontal axis. In order to determine the angle theta between the heading and the actual direction, we must subtract it from the given angle a, which is the direction of the wind vector with respect to the horizontal axis. Hence, the angle theta = a - (-9.7) = 66 degrees - (-9.7) = 75.7 degrees. Therefore, the angle theta between the heading and the actual direction of the airplane is approximately 75.7 degrees.