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One dimension kinematics, what am I doing wrong?

  • Thread starter hey123a
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  • #1
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Homework Statement


Car A is located 43km away, 17 degrees north of west. Car B is 34km away, 38 degrees east of north. The person in car A checks the position of car B by using a GPS. What does it give for car B's
a) distance from car 1
b) direction, measured due east?

Homework Equations


Pythagorean Theorem
Using sin, cos, tan


The Attempt at a Solution


Ax = (cos17)(43) = 41.12km
Ay = (sin17)(43) = 12.57km
Bx = (cos38)(43) = 26.79
By = (sin38)(34) = 20.93km

Cy = By - Ay
Cy = 20.93km - 12.57km
Cy = 8.36km

Cx = Bx + Ax
Cx = 26.79 + 41.12km
Cx = 67.91km

Vector C = sqrt[(Cy^2 + Cx^2)]
Vector C = sqrt[(8.36^2 + 67.91^2)]
a) Vector C = 68.42km

Angle = tan(Cx/Cy)
Angle = 67.91/8.36
b) Angle = 82.48 degrees
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,425
46

Homework Statement


Car A is located 43km away, 17 degrees north of west. Car B is 34km away, 38 degrees east of north. The person in car A checks the position of car B by using a GPS. What does it give for car B's
a) distance from car 1
b) direction, measured due east?

Homework Equations


Pythagorean Theorem
Using sin, cos, tan


The Attempt at a Solution


Ax = (cos17)(43) = 41.12km
Ay = (sin17)(43) = 12.57km
Bx = (cos38)(43) = 26.79
By = (sin38)(34) = 20.93km


Cy = By - Ay
Cy = 20.93km - 12.57km
Cy = 8.36km

Cx = Bx + Ax
Cx = 26.79 + 41.12km
Cx = 67.91km

Vector C = sqrt[(Cy^2 + Cx^2)]
Vector C = sqrt[(8.36^2 + 67.91^2)]
a) Vector C = 68.42km

Angle = tan(Cx/Cy)
Angle = 67.91/8.36
b) Angle = 82.48 degrees
You have the components of B back to front.

It is not good enough to just use cos for the x component and sin for the y component.
The 38 angle is from the vertical, North, where as the 34 angle was from the horizontal, West.
 
  • #3
37
0
You have the components of B back to front.

It is not good enough to just use cos for the x component and sin for the y component.
The 38 angle is from the vertical, North, where as the 34 angle was from the horizontal, West.
I don't get it what you're trying to say
 
  • #4
PeterO
Homework Helper
2,425
46
I don't get it what you're trying to say
Draw a sketch of the situation - making sure that your sketch versions of 34 degrees and 38 degress are actually smaller than 45 degrees.

For 34 degrees North of west, it will be obvious that the x component is bigger than the y component.

For the 38 degrees east of North, it should be equally obvious that the x component is smaller that the y component.

But in your calculations, you have Bx > By.

Alternatively

In the triangle formed for the "34 North of West" location, the y component is opposite the angle - so use sine, as you did - while the x component is adjacent to the angle - so use cosine, as you did.
In the triangle formed for the "38 East of North" location, the y component is adjacent to the angle - so use cosine, but you used sine - while the x component is opposite the angle - so use sine, but you used cosine.
 
  • #5
37
0
Draw a sketch of the situation - making sure that your sketch versions of 34 degrees and 38 degress are actually smaller than 45 degrees.

For 34 degrees North of west, it will be obvious that the x component is bigger than the y component.

For the 38 degrees east of North, it should be equally obvious that the x component is smaller that the y component.

But in your calculations, you have Bx > By.

Alternatively

In the triangle formed for the "34 North of West" location, the y component is opposite the angle - so use sine, as you did - while the x component is adjacent to the angle - so use cosine, as you did.
In the triangle formed for the "38 East of North" location, the y component is adjacent to the angle - so use cosine, but you used sine - while the x component is opposite the angle - so use sine, but you used cosine.
Oh east of north, so what i did in my calculation was north of east?
would i still get a right answer if did 90-38 = 52 degrees to have the y component opposite to the angle
 
  • #6
PeterO
Homework Helper
2,425
46
Oh east of north, so what i did in my calculation was north of east?
would i still get a right answer if did 90-38 = 52 degrees to have the y component opposite to the angle
Yes you would - but why would you bother?

Why your strong desire for the y component to be a sine - eventually (like here) you are going to get it backwards and get it wrong.

My approach is to work out both components - you often need them both anyway - then do a quick sketch - ensuring that if an angle is less than 45 degrees make sure it is sketched as such (better to make it too small than too big.

From the sketch you can see whether you want the smaller or larger component, and so just select the smaller or larger one as appropriate.

ie. don't spend time deciding whether you should be using sin or cos, concentrate on whether you want the small component or the large component.

example: a projectile is fired at 200m/s at an angle of 60 degrees to the horizontal - what is the vertical component of its initial velocity.
ans: the two components are 200.sin60 and 200.cos60 or 173.2m/s and 100 m/s
A quick sketch (perhaps just an imagined sketch) shows that the projectile is travelling faster up than horizontally, so the vertical component must be 173.2 m/s

compare that to: a cannonball is fired at 200m/s at an angle of 60 degrees to the vertical - what is the vertical component of its initial velocity.
ans: the two components are 200.sin60 and 200.cos60 or 173.2m/s and 100 m/s
A quick sketch (perhaps just an imagined sketch) shows that the projectile is travelling faster horizontally than vertically, so the vertical component must be 100 m/s
 
  • #7
PeterO
Homework Helper
2,425
46
Draw a sketch of the situation - making sure that your sketch versions of 34 degrees and 38 degress are actually smaller than 45 degrees.

For 34 degrees North of west, it will be obvious that the x component is bigger than the y component.

For the 38 degrees east of North, it should be equally obvious that the x component is smaller that the y component.

But in your calculations, you have Bx > By.

Note: when drawing the sketches, the key is whether the angles involved are smaller or larger than 45 degrees (at 45 degrees the components are equal)

In my sketches, I would draw both 34 and 38 degrees very small - like a real 10-15 degrees to exaggerate, for me, that one component is smaller than the other. The last thing you want is for you sketch, with its wonky lines, showing the components approximately the same size.
 
  • #8
37
0
Yes you would - but why would you bother?

Why your strong desire for the y component to be a sine - eventually (like here) you are going to get it backwards and get it wrong.

My approach is to work out both components - you often need them both anyway - then do a quick sketch - ensuring that if an angle is less than 45 degrees make sure it is sketched as such (better to make it too small than too big.

From the sketch you can see whether you want the smaller or larger component, and so just select the smaller or larger one as appropriate.

ie. don't spend time deciding whether you should be using sin or cos, concentrate on whether you want the small component or the large component.

example: a projectile is fired at 200m/s at an angle of 60 degrees to the horizontal - what is the vertical component of its initial velocity.
ans: the two components are 200.sin60 and 200.cos60 or 173.2m/s and 100 m/s
A quick sketch (perhaps just an imagined sketch) shows that the projectile is travelling faster up than horizontally, so the vertical component must be 173.2 m/s

compare that to: a cannonball is fired at 200m/s at an angle of 60 degrees to the vertical - what is the vertical component of its initial velocity.
ans: the two components are 200.sin60 and 200.cos60 or 173.2m/s and 100 m/s
A quick sketch (perhaps just an imagined sketch) shows that the projectile is travelling faster horizontally than vertically, so the vertical component must be 100 m/s
Ah okay well thank you very much! what got me confused was the directions, such as "north of east" or "east of north" and thats what mixed me up! i got the right answer now and well anyways thank you so much i appreciate you helping me out. the sketches definitely help out A LOT
Anyways I pressed the "thanks" button thank you so so much
 
  • #9
PeterO
Homework Helper
2,425
46
Ah okay well thank you very much! what got me confused was the directions, such as "north of east" or "east of north" and thats what mixed me up! i got the right answer now and well anyways thank you so much i appreciate you helping me out. the sketches definitely help out A LOT
Anyways I pressed the "thanks" button thank you so so much
happy to help - and good to see you (now?) see the value of a simple sketch.
 

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