How Do Displacement and Distance Differ in Particle Motion Calculations?

[xzibition8612]

Homework Statement

A particle's position is described by x=2t^3-4t-5
Find its a) displacement b) distance
from t=0 to t=5

Homework Equations

The Attempt at a Solution

Displacement, plug in x(5)-x(0) = 230m
Distance, shouldn't it be the same thing plug in and get 230m? But book says 234m. No idea how this came about.

 

[emailanmol]

Hey ,

Do you know the difference between distance and displacement (Am sure you do! :-))

If a particle goes from x = 2m to x=4m and then comes back to x=3m

What is the distance traveled and displacement?
How did you calculate it in simple terms.
Something similar is happening here.

 

[xzibition8612]

distance would be the actual physical length, so that would be 3m. Displacement is just the direct distance from start to finish, so that would be 1m.

 

[emailanmol]

Yes.So the basic logic you used is that you added the path length.
When the body reversed its direction at 4m ( in physics what is this point signify in terms of variables?) you didn't add the path from 4m to 3m as -1m.You added it as 1m.

Do the same on this question and you will get the answer:-)

 

[xzibition8612]

alright man but ur concept was a simple question, this one has complicated equations and i don't know how to apply it : ======================== (

 

[emailanmol]

Ok.
See displacement is integration of (vdt).
Distance is integration of (|v|*dt)
And |v| is speed.

For solving the mod || you need to find intervals where v is positive and where v is negative.

 

[xzibition8612]

so take absolute value of x=2t^3-4t-5.

then it would be -2t^3+45+5
so i take the integral of this and that's it?

 

[emailanmol]

For what value of t is the velocity positive?
When is it negative ?


take the negative of v where velocity is negative.

Take positive when velocity is positive.
Then integrate.
(Why do we do so?)

 

[xzibition8612]

ok, so when x=0 is the difference. So x is negative the function is: -2t^3+4t+5, and when x is positive the function is: 2t^3-4t-5

add these two up and integrate?
but if i add it up i get 0?

 

[emailanmol]

ok, so when x=0 is the difference. So x is negative the function is: -2t^3+4t+5, and when x is positive the function is: 2t^3-4t-5

add these two up and integrate?
but if i add it up i get 0?

No.We are interested in mod of velocity not x.take function as it is when v is positive(not neccesarily x) and multiply by minus when v is negative (not x)

 

[emailanmol]

ok, so when x=0 is the difference. So x is negative the function is: -2t^3+4t+5, and when x is positive the function is: 2t^3-4t-5

add these two up and integrate?
but if i add it up i get 0?

No.We are interested in mod of velocity not x.take function as it is when v is positive(not neccesarily x) and multiply by minus when v is negative (not x).


And you won't get 0.

 

[xzibition8612]

hey can you just show me the process? I'm getting very confused and I'm not sure what's going on. I think I can understand better if you show me the math. velocity would be x'=6t^2-4

Thanks man.

 

[emailanmol]

hey can you just show me the process? I'm getting very confused and I'm not sure what's going on. I think I can understand better if you show me the math. velocity would be x'=6t^2-4

Thanks man.

You calculated the velocity correctly. :-)

Property of mod function is that
when f(x) is positive

|f(x)| =f(x)

when f(x) is negative

|f(x)| =-f(x).

For eg if |2| is 2.|-2| is -(-2) i.e 2 again.

------------------

Velocity is 6t^2-4.
which is positive when t belongs from sqroot (2/3) to 5 and negative between 0 to sqroot(2/3.)
So take displacement of [-(2t^3-4t-5)] for t = 0 to t=sqroot(2/3).
(or you can integrate
[-(3t^2-4)dt]


And take displacement of [(2t^3-4t-5)] for t=sqroot(2/3) to t=5.
(or you can integrate
[(3t^2-4)dt]

 

[xzibition8612]

how did you get sqrt(2/3)??
I guess that's from x=2t^3-4t-5??
so 0=2t^3-4t-5
But how do I solve this cubic equation?

Thanks a lot man.

 

[emailanmol]

From 6t^2 -4 >0
We are interested in |v| not mod x.

See in the example i gave you .

From 2 to 4 m velocity was positive so you did 4-2 =2
(final position -initial position)

Then particle turned behind and velocity became negative.in moving from 4 to 3 you therefore did -(3-4)=1

And added these.

All this while x was positive.

Our sense of mod depends on velocity and not position :-)

My pleasure :-)

I hope am making stuffs clear enough :-)
Hopefully

 

[xzibition8612]

thanks a lot man europeans are superior

 

[emailanmol]

Lol. I am an Indian :-)