Question

HELP!!! Consider the equations

I. v = v0 + ax
II. y = (2m) cos(kx), where k = 2m^-1
III. v^2 = ax

Which ones are dimensionally correct?

Answer 1

Well, I haven't heard of "dimensionally correct," but you can check the units to find which ones \(can't\) be true!

Answer 2

How? I am so confused with this problem!

Answer 3

This are dimensions:

Distance (x) = [L]
Area = [L^2]
Volume = [L^3]
Velocity (v) = [L]/[T]
Acceleration (a) = [L]/[T^2]
Energy = [M][L^2]/[T^2]

Answer 4

These are the dimensions:*

Answer 5

It is confusing, because I don't know what this is! Sorry!

Answer 6

I'm researching now.

Answer 7

Thank you!

Answer 8

Dimensionally correct is just saying that the units work out correctly.

Answer 9

?

Answer 10

I think we should skip that, since I still have two more questions, which I think are easier to deal with.

Answer 11

While John is traveling along a straight interstate highway, he notices that the mile marker reads 257 km. John travels until he reaches the 149 km marker and then retraces his path to the 177 km marker. What is John's resultant displacement from the 257 km marker? Answer in units of km.

Answer 12

Before you move on, just plug in the dimensions for each of the 3 equations (ie: v = L/T so put L/T in place of v...and so on). Then simplify the results and see which one is a valid equation...being that both sides are equivalent.

Answer 13

For John's trip, draw it out first:

Answer 14

That's what I've been doing, and I know that the first one isn't correct, but I don't know about the second and third one.

Answer 15

So what's his total displacement?

Answer 16

Should I subtract 257 - 177 to find his resultant displacement?

Answer 17

Exactly.

Answer 18

So, will his resultant displacement from the 257 km marker will be 80 km?

Answer 19

So, will his resultant displacement from the 257 km marker be 80 km?*

Answer 20

Yes.

Going back to the other one:\[v^2 = ax\]Plug in values to get:\[(\frac{L}{T})^2=\frac{L}{T^2}*L\]So that's equivalent.

Answer 21

Just do the same thing with the other 2 equations and see how it turns out.

Answer 22

So, the third one is dimensionally correct, right?

Answer 23

Yes...and the first one is not equivalent. Now try the second one

Answer 24

I'm pretty sure the second one is dimensionally correct.

Answer 25

How did you come to that conclusion?

Answer 26

Before you can answer that you need to know how y is defined. Do you have that info?

Answer 27

I actually did some research over the internet and found is dimensionally correct.

Answer 28

I'm very confused with the second one, since it is y = (2m) cos(kx), where k=2m^-1.

Answer 29

But you can figure this out on your own....it's very easy but I don't know what y is supposed to represent since it's not in your Dimensions post.

Answer 30

That's why I'm confused.

Answer 31

Here's a hint, the cosine of any value will be in what range?

Answer 32

-1 to 1.

Answer 33

So no matter what is inside those parenthesis, the result will be from 1 to -1...with no units. Therefore, you can actually forget about the cosine part of the equation altogether. Then you're left with this:\[y = 2m\]If y's units are the same as m's units, then the equation is valid.

Answer 34

But, what are the units for y

Answer 35

?

Answer 36

I have no idea...it's not in the question. If you found it on the internet, maybe that question actually has more info in it.

Answer 37

If y is something like the typical vertical distance (meters), then the equation would be considered dimensionally correct since the units on both sides are equal.

Answer 38

That's probably a safe assumption in this case but I'd argue that the question is missing info.

Answer 39

Thank you! Now, going back to the displacement question, I'm not sure about the answer I got.

Answer 40

What isn't clear to you in that one? Think about the definition of displacement.

Answer 41

I just find it too simple to be truth.

Answer 42

I mean, 257-177 = 80, and 80 is the final answer?

Answer 43

The question is intended to test your understanding of displacement versus distance...that's all. Yes, the final answer is 80km.

Answer 44

But, what happened with the 149 km he reaches before he retraced his displacement to the 177 km marker.

Answer 45

?

Answer 46

But, what happened with the 149 km he reaches before he retraced his path to the 177 km marker?*

Answer 47

It doesn't count in displacement. You can think of displacement as the direct-line distance between two points. In the end, his displacement from the 257KM marker was only 80KM...it doesn't matter what path he took to get there.

If the question asked for distance, it would be a different story.

Answer 48

When I enter the answer 80 km (since it is an online homework system), the system says that it is incorrect.