- #1

quicksilver123

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## 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.

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