# Determine the particles position

1. Jan 28, 2010

### patflo

1. The acceleration of a particle along a straight line is defined by a= (2t-9)m/s² where t is in seconds. At t=0,s=1m and v=10m/s. When t=9s determine a) the particles position b) the total distance travelled and c) the velocity

2. I think: a=dv/dt; v= ds/dt

3. From a=dv/dt you integrate and get v; from v=ds/dt you integrate and get s
Then you put in for t=9s and get the result. but probably i failed the integration, what is the value of the constant you get from the integrations?

2. Jan 29, 2010

Could you show us your work on this one?

3. Jan 30, 2010

### patflo

v=integrate(a dt)=t²-9t+v0 t=0 therefore vo=10
v=t²-9t+10

s=inegrate(v dt) =t³/3-9t²/2+10t+s0 t=0 s=1 therefore s0=1

s=t³/3-9t²/2+10t+1

is that correct?

4. Jan 30, 2010

### jdwood983

The integrations do work out to what you have, so now you must use t=9s to find the position and velocity.

5. Feb 1, 2010

### patflo

ok..now i got for position after 9s: s(9)=-30,5m that means the car travelled to the left if i suppose that the positive direction is to the right.

Same thing with velocity v(9)= 10 m/s

But whats about the total distance travelled? how can i solve that? if i put all the values from 0 to 9s in the equation for s then i see that first the car moves to the right than to the left and then it stops at -30,5m. please help me?

6. Feb 1, 2010

### jdwood983

If you start at x=1m and then move to -30.5 m, how far have you traveled?

7. Feb 1, 2010

### patflo

yeah i know..but try to put s(1,5)=7m, s(8)=-36m

so that would mean that i go 7m to the right, than 7m back (in total 14m) than -36m (so 50m total) and then to -30,5.

total distance travelled approxiamtely= 55,5m

but i have no analytiv solving for this problem...its only by trying out some points of time.

8. Feb 2, 2010

### jdwood983

I think Post #6 holds your answer and that you are thinking a little too hard about this part of the problem. Unless I am mistaken as to the wording of the question, I read that as total displacement: difference between starting point and ending point which will be 31,5 m.

9. Feb 2, 2010

### gabbagabbahey

I disagree with jdwood's interpretation. To me, "distance traveled" and "displacement" are not the same thing. If I drive around the block a few times and end up at the exact same spot I started, my displacement will be zero, but my odometer will confirm that I traveled a certain non-zero distance.

Mathematically, displacement is the definite integral of velocity w.r.t time, while distance traveled is the definite integral of speed w.r.t time.

Since the motion in this case is one-dimensional, the only thing that distinguishes speed from velocity is the sign of the velocity....just take its absolute value and integrate it from t=0 to t=9...and you will get the distance traveled.

10. Feb 2, 2010

### patflo

i will try this out..thanks!..i i will inform you if it worked!!