Simple Relative Motion Problem

In summary, the velocity of the ladybug, relative to the ground, is 47.1mm/s [W 40.6deg N]. The calculations involve using component vectors and the Pythagorean theorem, and it is important to consider significant figures in the final answer.
  • #1
quicksilver123
173
0

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 = VLC = 10mm/s [W]
Components:
VLCx = 10mm/s
VLCy = 0mm/s

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

Velocity of the ladybug, relative to the ground = VLG = ?
Components:
VLGx = VLCx + VCGx = 35.71150439mm/s
VLGy = VLCy + VCGy = 30.64177772mm/s

Direction of VLG
∅ (theta?) = tan-1(opposite/adjacent)
= tan-1(VLGy / VLGx)
= tan-1(30.64177772mm/s / 35.71150439mm/s)
= 40.63 degrees

∴ VLG = √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.
 
Last edited:
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  • #2
quicksilver123 said:

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 = VLC = 10mm/s [W]
Components:
VLCx = 10mm/s
VLCy = 0mm/s

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

Velocity of the ladybug, relative to the ground = VLG = ?
Components:
VLGx = VLCx + VCGx = 35.71150439mm/s
VLGy = VLCy + VCGy = 30.64177772mm/s

Direction of VLG
∅ (theta?) = tan-1(opposite/adjacent)
= tan-1(VLGy / VLGx)
= tan-1(30.64177772mm/s / 35.71150439mm/s)
= 40.63 degrees

∴ VLG = √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.
 
  • #3
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?
 
  • #4
Looks good.
 
  • #5
Thanks bros (and broettes).
 

Related to Simple Relative Motion Problem

What is a "Simple Relative Motion Problem"?

A "Simple Relative Motion Problem" is a type of problem in physics that involves calculating the motion of one object relative to another. This can include determining the speed, direction, and position of an object in relation to another object or a fixed point.

What are the key concepts involved in solving a "Simple Relative Motion Problem"?

The key concepts involved in solving a "Simple Relative Motion Problem" include understanding displacement, velocity, and acceleration, as well as the concept of frames of reference. It is also important to have a solid grasp of vector addition and subtraction, as these are often used to solve such problems.

How do you determine the frame of reference in a "Simple Relative Motion Problem"?

The frame of reference in a "Simple Relative Motion Problem" is typically determined by identifying a fixed point or object that will be used as a reference point for all measurements. This can be a stationary object or a point on the coordinate system. It is important to establish a consistent frame of reference to accurately calculate the motion of the objects involved.

What are some common strategies for solving "Simple Relative Motion Problems"?

Some common strategies for solving "Simple Relative Motion Problems" include breaking down the problem into smaller parts, drawing diagrams to visualize the motion, and using equations such as the distance formula and the equations of motion. It can also be helpful to identify and label the known and unknown variables in the problem.

What are some real-world applications of "Simple Relative Motion Problems"?

"Simple Relative Motion Problems" have many real-world applications, including determining the motion of vehicles on a road, calculating the trajectory of projectiles, and predicting the movement of celestial bodies in space. These problems are also important in fields such as engineering, navigation, and sports science.

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