Asked by Anonymous
Can you please help me correct my answers for the following two questions?
1) A tour boat travels 25 km due east and then 15 km S50°E. Represent these displacements in a vector diagram, then calculate the resultant displacement.
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My Work:
I drew the vectors and connected them head to tail to form a triangle.
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Then, I found the resultant displacement:
|r|² = 25² + 15² - 2(15)(25)cos130
√|r|² = √1332.1
|r| = 36.5 km
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Now, I'm having some problems finding the direction of the resultant displacement:
(sin50/36.5) = (sinC/25)
31.6° = C
So, I get 36.5 km S31.6°E
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The textbook answer is 36.5 km S54°E
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2) Vectors a and b have magnitudes 2 and 3, respectively. If the angle between them is 50°, find the vectors "5a - 2b", and state its magnitude and direction.
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Here's my work for this question:
I drew the vectors and connected them head to tail.
-----
I tried to find the magnitude by:
5a - 2b
= 5(2a) - 2(3b)
= 10a - 6b
|r|² = 10² + 6² - 2(10)(6)cos130
√|r|² = √213
|r| = 14.6
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Then I tried to find the direction by:
(sinx/3) = (sin130/14.6)
x = 9.1°
So, I get 14.6, 9.1° to vector a
======
Textbook answer is 7.7, 37° to vector a
1) A tour boat travels 25 km due east and then 15 km S50°E. Represent these displacements in a vector diagram, then calculate the resultant displacement.
======
My Work:
I drew the vectors and connected them head to tail to form a triangle.
-----
Then, I found the resultant displacement:
|r|² = 25² + 15² - 2(15)(25)cos130
√|r|² = √1332.1
|r| = 36.5 km
-----
Now, I'm having some problems finding the direction of the resultant displacement:
(sin50/36.5) = (sinC/25)
31.6° = C
So, I get 36.5 km S31.6°E
======
The textbook answer is 36.5 km S54°E
_____________________________________
2) Vectors a and b have magnitudes 2 and 3, respectively. If the angle between them is 50°, find the vectors "5a - 2b", and state its magnitude and direction.
======
Here's my work for this question:
I drew the vectors and connected them head to tail.
-----
I tried to find the magnitude by:
5a - 2b
= 5(2a) - 2(3b)
= 10a - 6b
|r|² = 10² + 6² - 2(10)(6)cos130
√|r|² = √213
|r| = 14.6
-----
Then I tried to find the direction by:
(sinx/3) = (sin130/14.6)
x = 9.1°
So, I get 14.6, 9.1° to vector a
======
Textbook answer is 7.7, 37° to vector a
Answers
Answered by
Reiny
I will do the second question first, since I have a question about your interpretation of S50E from the first.
I drew the 2 unit vector to run east, then the 3 unit angle downwards to form the 50 degree angle.
so when you construct 5a - 2b, you would draw a horizontal line 10 units long for the first part, then you must go into the opposite direction of the second vector for 6 units
so the magnitude equation would be
|r|² = 10² + 6² - 2(10)(6)cos50
and r = 7.67
now if x is the angle between the resultant and the first vector
sinx/6 = sin50/7.67
I got x = 36.8 degrees
back to your first problem,
I was always under the impression and taught that a direction like your
S50°E meant:
face south then 50 degrees towards the east, so I thought that the end part of your equation should have been ....cos140
you had....cos130
but the ....cos130 produced the answer supplied by your text, so I am confused.
I am using the Canadian interpretation of S50°E, is it different where you are???
Perhaps some of the other math or physics expert could help out here
I drew the 2 unit vector to run east, then the 3 unit angle downwards to form the 50 degree angle.
so when you construct 5a - 2b, you would draw a horizontal line 10 units long for the first part, then you must go into the opposite direction of the second vector for 6 units
so the magnitude equation would be
|r|² = 10² + 6² - 2(10)(6)cos50
and r = 7.67
now if x is the angle between the resultant and the first vector
sinx/6 = sin50/7.67
I got x = 36.8 degrees
back to your first problem,
I was always under the impression and taught that a direction like your
S50°E meant:
face south then 50 degrees towards the east, so I thought that the end part of your equation should have been ....cos140
you had....cos130
but the ....cos130 produced the answer supplied by your text, so I am confused.
I am using the Canadian interpretation of S50°E, is it different where you are???
Perhaps some of the other math or physics expert could help out here
Answered by
Nyamekye
a body is acting upon by two force 8 newton in the direction 60degree north east.find the magnitude and the direction of the reasultant force on the body using parallel gram of vectors
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