Simple Relative Motion Problem

[quicksilver123]

Homework Statement

A ladybug with a velocity of 10.0mm/s [W] crawls on a chair that is being pulled [50deg N of W] at 40.0mm/s. What is the velocity of the ladybug, relative to the ground?

Homework Equations

pythag theorum
component vectors
sohcahtoa

The Attempt at a Solution

Let [N] and [W] be positive.

Velocity of the ladybug relative to the chair = V_LC = 10mm/s [W]
Components:
V_LCx = 10mm/s
V_LCy = 0mm/s

Velocity of the chair, relative to the ground = V_CG = 40mm/s [W50degN]
Components:
V_CGx = 40mm/s (cos50) = 25.71150439mm/s
V_CGy = 40mm/s (sin50) = 30.64177772mm/s

Velocity of the ladybug, relative to the ground = V_LG = ?
Components:
V_LGx = V_LCx + V_CGx = 35.71150439mm/s
V_LGy = V_LCy + V_CGy = 30.64177772mm/s

Direction of V_LG
∅ (theta?) = tan^-1(opposite/adjacent)
= tan^-1(V_LGy / V_LGx)
= tan^-1(30.64177772mm/s / 35.71150439mm/s)
= 40.63 degrees

∴ V_LG = √2214.230088 = 47.05560634mm/s [ W 40.63deg N ]I feel that my calculations are correct.
However, looking up this question on the net shows that people have come to a different end result:
https://www.physicsforums.com/showthread.php?t=596229

Specifically, the direction of the resultant vector is different.
I just want to be sure that my answer is correct before I submit it for marking.
Thanks.

 

[gneill]

Homework Statement

A ladybug with a velocity of 10mm/s [W] crawls on a chair that is being pulled [50deg N of W] at 40.0mm/s. What is the velocity of the ladybug, relative to the ground?

Homework Equations

pythag theorum
component vectors
sohcahtoa

The Attempt at a Solution

Let [N] and [W] be positive.

Velocity of the ladybug relative to the chair = V_LC = 10mm/s [W]
Components:
V_LCx = 10mm/s
V_LCy = 0mm/s

Velocity of the chair, relative to the ground = V_CG = 40mm/s [W50degN]
Components:
V_CGx = 40mm/s (cos50) = 25.71150439mm/s
V_CGy = 40mm/s (sin50) = 30.64177772mm/s

Velocity of the ladybug, relative to the ground = V_LG = ?
Components:
V_LGx = V_LCx + V_CGx = 35.71150439mm/s
V_LGy = V_LCy + V_CGy = 30.64177772mm/s

Direction of V_LG
∅ (theta?) = tan^-1(opposite/adjacent)
= tan^-1(V_LGy / V_LGx)
= tan^-1(30.64177772mm/s / 35.71150439mm/s)
= 40.63 degrees

∴ V_LG = √2214.230088 = 47.05560634mm/s [ W 40.63deg N ]


I feel that my calculations are correct.
However, looking up this question on the net shows that people have come to a different end result:
https://www.physicsforums.com/showthread.php?t=596229

Specifically, the direction of the resultant vector is different.



I just want to be sure that my answer is correct before I submit it for marking.
Thanks.

The result of your calculations look okay. You might want to do something about significant figures for the final value before handing it in.

 

[quicksilver123]

I haven't really done much math this year. From what I recall, you use the given values in determining the amount of significant digits... you choose the number that has the least amount of sig. digits.

Both 40.0mm/s and 10.0mm/s have three sig digits.

So...
The final, rounded answer should be:

VLG = = 47.1mm/s [ W 40.6deg N ]

Correct?

 

[gneill]

Looks good.

 

[quicksilver123]

Thanks bros (and broettes).