Mixing
Displacement vectors
$begingroup$
You walk $53$ m to the north, then turn $60$ degrees to your right and walk another $45$ m. Determine the direction of your displacement vector. Express your answer as an angle relative to east.
So I did pythagorean theorem and got the adjacent side to be $28$. Then I did inverse tan and got the angle to be $50$ degrees. Help please?
The answer is $63$ degrees.
physics
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
asked Oct 15 '14 at 22:28
LilLil
95532342
$endgroup$
add a comment |
$begingroup$
You walk $53$ m to the north, then turn $60$ degrees to your right and walk another $45$ m. Determine the direction of your displacement vector. Express your answer as an angle relative to east.
So I did pythagorean theorem and got the adjacent side to be $28$. Then I did inverse tan and got the angle to be $50$ degrees. Help please?
The answer is $63$ degrees.
physics
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
asked Oct 15 '14 at 22:28
LilLil
95532342
$endgroup$
$begingroup$
for clarification, we are assuming euclidean geometry? In otherwords, is our map a flat grid?
$endgroup$
– JMoravitz
Oct 15 '14 at 22:31
$begingroup$
yes. This is actually a beginners physics problem
$endgroup$
– Lil
Oct 15 '14 at 22:32
add a comment |
$begingroup$
You walk $53$ m to the north, then turn $60$ degrees to your right and walk another $45$ m. Determine the direction of your displacement vector. Express your answer as an angle relative to east.
So I did pythagorean theorem and got the adjacent side to be $28$. Then I did inverse tan and got the angle to be $50$ degrees. Help please?
The answer is $63$ degrees.
physics
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
asked Oct 15 '14 at 22:28
LilLil
95532342
$endgroup$
You walk $53$ m to the north, then turn $60$ degrees to your right and walk another $45$ m. Determine the direction of your displacement vector. Express your answer as an angle relative to east.
So I did pythagorean theorem and got the adjacent side to be $28$. Then I did inverse tan and got the angle to be $50$ degrees. Help please?
The answer is $63$ degrees.
physics
physics
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
asked Oct 15 '14 at 22:28
LilLil
95532342
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
asked Oct 15 '14 at 22:28
LilLil
95532342
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
edited Feb 10 '16 at 2:32
JKnecht
5,12651631
5,12651631
asked Oct 15 '14 at 22:28
LilLil
95532342
asked Oct 15 '14 at 22:28
LilLil
95532342
asked Oct 15 '14 at 22:28
LilLil
95532342
95532342
$begingroup$
for clarification, we are assuming euclidean geometry? In otherwords, is our map a flat grid?
$endgroup$
– JMoravitz
Oct 15 '14 at 22:31
$begingroup$
yes. This is actually a beginners physics problem
$endgroup$
– Lil
Oct 15 '14 at 22:32
add a comment |
$begingroup$
for clarification, we are assuming euclidean geometry? In otherwords, is our map a flat grid?
$endgroup$
– JMoravitz
Oct 15 '14 at 22:31
$begingroup$
yes. This is actually a beginners physics problem
$endgroup$
– Lil
Oct 15 '14 at 22:32
$begingroup$
for clarification, we are assuming euclidean geometry? In otherwords, is our map a flat grid?
$endgroup$
– JMoravitz
Oct 15 '14 at 22:31
$begingroup$
for clarification, we are assuming euclidean geometry? In otherwords, is our map a flat grid?
$endgroup$
– JMoravitz
Oct 15 '14 at 22:31
$begingroup$
yes. This is actually a beginners physics problem
$endgroup$
– Lil
Oct 15 '14 at 22:32
$begingroup$
yes. This is actually a beginners physics problem
$endgroup$
– Lil
Oct 15 '14 at 22:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45sin alpha, 53 + 45 cos alpha)$. $sin alpha = frac{sqrt 3}{2}, cos alpha = frac 12$, so the resulting point is $(frac{45sqrt3}{2},frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
$endgroup$
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
add a comment |
$begingroup$
If you were to draw a graph, the first leg of the trip would go straight up, and once you reach the second leg of the trip you can recognize that a triangle forms above it, looking like a flag pole with a triangular flag at the top. Using your properties of that smaller triangle, you can now look at the bigger triangle and solve from there.
From here you can find the total height of the larger triangle to be $53+45cdot frac{1}{2} = 75.5$ and the width to be $frac{45sqrt{3}}{2}$.
Using Inverse tangent on these will give you the angle at the bottom corner of the triangle. Since it asks for angle relative to east however, you'll want to take 90 degrees minus that to get the complement.
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45sin alpha, 53 + 45 cos alpha)$. $sin alpha = frac{sqrt 3}{2}, cos alpha = frac 12$, so the resulting point is $(frac{45sqrt3}{2},frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
$endgroup$
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
add a comment |
$begingroup$
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45sin alpha, 53 + 45 cos alpha)$. $sin alpha = frac{sqrt 3}{2}, cos alpha = frac 12$, so the resulting point is $(frac{45sqrt3}{2},frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
$endgroup$
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
add a comment |
$begingroup$
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45sin alpha, 53 + 45 cos alpha)$. $sin alpha = frac{sqrt 3}{2}, cos alpha = frac 12$, so the resulting point is $(frac{45sqrt3}{2},frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
$endgroup$
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45sin alpha, 53 + 45 cos alpha)$. $sin alpha = frac{sqrt 3}{2}, cos alpha = frac 12$, so the resulting point is $(frac{45sqrt3}{2},frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
answered Oct 15 '14 at 22:36
Andrei RykhalskiAndrei Rykhalski
1,212614
1,212614
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
add a comment |
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
is there a way to solve this using the pythagorean theorem and SOH-CAH-TOA?
$endgroup$
– Lil
Oct 15 '14 at 22:37
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
$begingroup$
Yes, on step 2 you may construct a right triangle with hypotenuse = 45 and angle of 60 degrees and find x and y components of displacement between (0,53) and the final point from that triangle.
$endgroup$
– Andrei Rykhalski
Oct 15 '14 at 22:44
add a comment |
$begingroup$
If you were to draw a graph, the first leg of the trip would go straight up, and once you reach the second leg of the trip you can recognize that a triangle forms above it, looking like a flag pole with a triangular flag at the top. Using your properties of that smaller triangle, you can now look at the bigger triangle and solve from there.
From here you can find the total height of the larger triangle to be $53+45cdot frac{1}{2} = 75.5$ and the width to be $frac{45sqrt{3}}{2}$.
Using Inverse tangent on these will give you the angle at the bottom corner of the triangle. Since it asks for angle relative to east however, you'll want to take 90 degrees minus that to get the complement.
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
$endgroup$
add a comment |
$begingroup$
If you were to draw a graph, the first leg of the trip would go straight up, and once you reach the second leg of the trip you can recognize that a triangle forms above it, looking like a flag pole with a triangular flag at the top. Using your properties of that smaller triangle, you can now look at the bigger triangle and solve from there.
From here you can find the total height of the larger triangle to be $53+45cdot frac{1}{2} = 75.5$ and the width to be $frac{45sqrt{3}}{2}$.
Using Inverse tangent on these will give you the angle at the bottom corner of the triangle. Since it asks for angle relative to east however, you'll want to take 90 degrees minus that to get the complement.
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
$endgroup$
add a comment |
$begingroup$
If you were to draw a graph, the first leg of the trip would go straight up, and once you reach the second leg of the trip you can recognize that a triangle forms above it, looking like a flag pole with a triangular flag at the top. Using your properties of that smaller triangle, you can now look at the bigger triangle and solve from there.
From here you can find the total height of the larger triangle to be $53+45cdot frac{1}{2} = 75.5$ and the width to be $frac{45sqrt{3}}{2}$.
Using Inverse tangent on these will give you the angle at the bottom corner of the triangle. Since it asks for angle relative to east however, you'll want to take 90 degrees minus that to get the complement.
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
$endgroup$
If you were to draw a graph, the first leg of the trip would go straight up, and once you reach the second leg of the trip you can recognize that a triangle forms above it, looking like a flag pole with a triangular flag at the top. Using your properties of that smaller triangle, you can now look at the bigger triangle and solve from there.
From here you can find the total height of the larger triangle to be $53+45cdot frac{1}{2} = 75.5$ and the width to be $frac{45sqrt{3}}{2}$.
Using Inverse tangent on these will give you the angle at the bottom corner of the triangle. Since it asks for angle relative to east however, you'll want to take 90 degrees minus that to get the complement.
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
answered Oct 15 '14 at 22:50
JMoravitzJMoravitz
46.6k33886
46.6k33886
add a comment |
add a comment |
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$begingroup$
for clarification, we are assuming euclidean geometry? In otherwords, is our map a flat grid?
$endgroup$
– JMoravitz
Oct 15 '14 at 22:31
$begingroup$
yes. This is actually a beginners physics problem
$endgroup$
– Lil
Oct 15 '14 at 22:32