Dimensional motion

1. Sep 16, 2007

innightmare

1. The problem statement, all variables and given/known data
A motorists drives south at 20.0 m/s for 3.0min, then turns west and travels at 25.0m/s for 2.0min and finally travels northwest at 30.0m/s for 1.00min. For this 6.00 min trip, find:
a)the total vector displacement
b) the average speed
c)the average velocity
Let the positive x-axis point east

2. Relevant equations

v=deltavectorR/delta-t
vector-v=d-vectorR/dt=dx/dt(i)+dy/dt(j)
vectorR=vectorR(i)+vector-velocity(t)+1/2vector-at^2

3. The attempt at a solution
I drew a diagram, to make the problem clearer. I think t=6min xi=25.om/s. But then I have two y-components

2. Sep 16, 2007

learningphysics

How did you approach part a) ?

3. Sep 16, 2007

innightmare

its displacement vector: deltavectorR=vectorR(f)-vectorR(i)

But I dont know how to go about the final and initial displacement. I am confused

4. Sep 17, 2007

learningphysics

Find the displacement in the east/west direction... then the north/south direction...

I'd write each displacement as a vector (taking i horizontally, j vertically, postive east and north):

ie: the first part (20*3*60)

$$\vec{r} = -3600\vec{j}$$

Does that make sense... same way write all 3 displacements. Then add them.

5. Sep 17, 2007

innightmare

ok, its starting to kick in. now for b): wouldnt average velocity be=my value for deltavectorr/delta t??

6. Sep 17, 2007

learningphysics

yes, that would be for average velocity... average speed will be different.

7. Sep 17, 2007

innightmare

OK-so if my displacement are in terms of vectors, then how in the world is b) computed. B/c the average velocity is deltaR/delta t...but my delta-displacement is going to be in vectors. how in the world am i supposed to divide vectors.

also average speed is total distance/delta t.

8. Sep 17, 2007

learningphysics

You need to find the total distance travelled. then divide by time.

9. Sep 17, 2007

innightmare

for the total distance do i add the total values given? and what about average velocity?

10. Sep 17, 2007

learningphysics

find the distance travelled in each segment and add them up. distance is not displacement. With distance, you're ignoring direction.

Average velocity is the total displacement/time.

Did you calculate the displacement vectors for each section. Add up those vectors and you get the total displacement.

For total distance, you're just adding up numbers, not vectors.

11. Sep 17, 2007

innightmare

for the average velocity, isnt my displacement in vectors, and if so then how is it divided by time? i did calculate the displacement for each section

12. Sep 17, 2007

learningphysics

Add up all the displacements of each section.... that gives total displacement as a vector... divide that by the total time... we can divide vectors by scalars... time is just a scalar.

13. Sep 18, 2007

innightmare

OK-Ive added my vectors, but it still hasnt given me the same answers in the book. In the book:a)4.87km at 209 degree from east
b) 23.3 ms
c)13.5m/s at 209
I converted the given values into meters. for instance:{3.0min*60s/min}which then i computed 20.0m/s*180s=-3600
This was what ive done for each one.
then (0,-vt)=(0,-3600)
(v2t2,-vt)= (3000, -3600)
(v3t3)=(-2545.6,3600)

This is the part where i get lost. for vector i=2745.4, j=-3600
If the question asks for total vector displacement, why is the answer 4.87 at an 209 angle

For b) my time would be 6min, but for my total length, would i total my values i converted into meters?

AGAIN-thank you sooooo much for your patience

14. Sep 18, 2007

learningphysics

I don't understand your numbers... Are you getting the displacement over each section? For example... 25.0m/s west for 2.0min is a displacement of (-3000,0). How do you get (3000,-3600) ?

15. Sep 19, 2007

innightmare

those are my x and y components. im over-complicated this, arent i