John and Mary are skating at an ice rink. John skates at a constant speed of 6.7 m/s, with respect to the ice surface, directly south. Mary skates at a constant speed of 11.5 m/s in a direction of 34° west of south, with respect to the ice. What are the magnitude and direction of Mary's velocity with respect to John? Give the direction as measured relative to due south.
I don't understand how to graph this question. SOMEONE PLEASE HELP.
Well, it sounds like John is skating directly south, so we can imagine him moving straight down the graph. Now, Mary is skating at an angle of 34° west of south, which means we need to turn her velocity arrow 34° clockwise from due south. Then, we just need to compare the lengths of John and Mary's velocity arrows to figure out the magnitude of Mary's velocity with respect to John.
Since we don't have specific values for the lengths of the velocity arrows, let's just call John's velocity arrow length "J" and Mary's velocity arrow length "M". Now, since John's velocity is constant at 6.7 m/s, we can draw a line straight down to represent his velocity arrow with length J.
For Mary, we need to break down her velocity into its south and west components. The south component of her velocity will be M * sin(34°), and the west component will be M * cos(34°).
To graph Mary's velocity arrow, we start at John's position and move M * sin(34°) units straight down to represent the south component, then move M * cos(34°) units to the right.
The arrow from John's position to Mary's position represents the magnitude and direction of Mary's velocity with respect to John. To find this magnitude, we can use the Pythagorean theorem since John and Mary's velocity arrows form a right triangle.
Now, I know this explanation may not be very helpful without a visual representation, but I hope it gives you a basic understanding of how to approach this graphing problem. Remember, if in doubt, always ask a math teacher or consult your textbook for more guidance. Good luck!
To solve this problem, it is helpful to break down the velocities of John and Mary into their north-south components.
John's velocity is directly south, so it only has a southward component. The magnitude of John's velocity is given as 6.7 m/s.
Mary's velocity is at an angle of 34° west of south. To find the north-south components of Mary's velocity, we can use trigonometry.
The southward component of Mary's velocity can be found using the cosine function:
Southward component = magnitude of velocity * cos(angle)
Southward component = 11.5 m/s * cos(34°)
Southward component ≈ 11.5 m/s * 0.829 = 9.537 m/s
To find the northward component of Mary's velocity, we can use the sine function:
Northward component = magnitude of velocity * sin(angle)
Northward component = 11.5 m/s * sin(34°)
Northward component ≈ 11.5 m/s * 0.560 = 6.44 m/s
Now that we have the northward and southward components of Mary's velocity, we can calculate the magnitude and direction of Mary's velocity with respect to John.
The magnitude of Mary's velocity with respect to John can be found using the Pythagorean theorem:
Magnitude = sqrt(southward component^2 + northward component^2)
Magnitude = sqrt(9.537^2 + 6.44^2) ≈ sqrt(90.806 + 41.5536) ≈ sqrt(132.3596) ≈ 11.51 m/s
The direction of Mary's velocity with respect to John can be found by determining the angle between Mary's velocity vector and due south.
Direction = atan(northward component / southward component)
Direction = atan(6.44 m/s / 9.537 m/s) ≈ atan(0.676) ≈ 33.45°
Therefore, the magnitude of Mary's velocity with respect to John is approximately 11.51 m/s, and the direction is approximately 33.45° west of south.
To graph this question, we can use a graphical method called vector addition. This method involves drawing vectors to scale and then adding them geometrically to find the resultant vector.
Let's start by drawing a coordinate system where the x-axis represents east-west direction and the y-axis represents north-south direction. The positive x-direction is east, and the positive y-direction is north.
Now, let's draw John's velocity vector. Since he skates directly south, we draw an arrow pointing downwards on the y-axis, with a length of 6.7 units (representing 6.7 m/s) directly below the origin.
Next, let's draw Mary's velocity vector. She skates in a direction that is 34° west of south. To do this, we draw an arrow with a length of 11.5 units (representing 11.5 m/s) at an angle of 34° counter-clockwise from the negative y-axis.
To find the magnitude and direction of Mary's velocity with respect to John, we need to find the vector sum of the two velocity vectors. We can do this by placing the tail of Mary's velocity vector at the head of John's velocity vector and then drawing a new vector from the tail of John's vector to the head of Mary's vector.
This new vector represents the resultant velocity, i.e., Mary's velocity with respect to John. We can measure its magnitude (length) using a ruler or a scale and measure its direction using a protractor or by referencing the angle from the positive y-axis.
Once we measure the magnitude and direction, we can then report them as the answer to the question.
Note: In case you find it difficult to graphically determine the magnitude and direction of the resultant vector, you can also solve it algebraically using trigonometry and vector addition formulas.
using vectors,
John's velocity is <0,-6.7>
Mary's velocity is <-6.43,-9.53>
Mary's velocity with respect to John is the difference between their velocities:
<-6.43,-9.53> - <0,-6.7> = <-6.43,-2.83>
Now just get the magnitude and direction of that vector:
7.03m/s at W23.7°S