IBV10 using displacement vectors

[karush]

https://www.physicsforums.com/attachments/1341

for (i) just did \(\displaystyle 16^2+12^2=400\) so \(\displaystyle \sqrt{400} = 20\) km

but don't know how they got the \(\displaystyle 13 \frac{km}{min}\)

there are some more ? on this involving the displacement vectors but want to get this (a) part done first.

 

[topsquark]

https://www.physicsforums.com/attachments/1341

for (i) just did \(\displaystyle 16^2+12^2=400\) so \(\displaystyle \sqrt{400} = 20\) km

but don't know how they got the \(\displaystyle 13 \frac{km}{min}\)

there are some more ? on this involving the displacement vectors but want to get this (a) part done first.

I can't think of a way to do this without Calculus. Well, here it is anyway.

\(\displaystyle v = \frac{dr}{dt}\)

So take the derivative of r, compute v, then find the "size" of v.

-Dan

 

[MarkFL]

I suppose one could assume that given a linear position function, the velocity must be constant...:D

 

[topsquark]

I suppose one could assume that given a linear position function, the velocity must be constant...:D

Nah. Too simple.

Good catch.

-Dan

 

[karush]

OK, if the velocity is constant.
then there is a slope, since \(\displaystyle t\pmatrix{12 \\ -5}\) is Time x Rate then\(\displaystyle \sqrt{(12)^2+(-5)^2}= 13\) km/min

now we have a point $(16, 12)$ and $m=-\frac{5}{12}$
from which we can derive the eq of
$5x+12y=224$

if correct, I did this by conjecture, not knowing the formal process.

still some more ? on this...

 

[topsquark]

now we have a point $(16, 12)$ and $m=-\frac{5}{12}$
from which we can derive the eq of
$5x+12y=224$

In the future, please post all of the question you need help with.

You seem to be trying to find the line between two points, one of which is (16, 12). What is the other point?

-Dan

 

[karush]

In the future, please post all of the question you need help with.

You seem to be trying to find the line between two points, one of which is (16, 12). What is the other point?

-Dan

ok, thot this was in the OP but it wasn't

The question was:
Show that the Cartesian equation of the path of Air One is:

$5x+12y=224$

However, from $r_1=\pmatrix{16 \\ 12} + t\pmatrix{12 \\ -5}$

we have point $(16, 12)$ and $m=-\frac{5}{12}$

we can derive $5x+12y=224$