Thursday

September 3, 2015
Posted by **Barbara** on Tuesday, January 6, 2009 at 9:28pm.

- 12th grade Subject?? -
**Ms. Sue**, Tuesday, January 6, 2009 at 9:30pmWhat is your subject?

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:31pmCalculus

- 12th grade -
**bobpursley**, Tuesday, January 6, 2009 at 9:31pmI assume you are in calculus.

Postition=INTEGRAL v(t) dt=INT (2t+1)dt

= t^2 + t

Put in t=4 and compute.

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:34pmso my answer would basically be 20

- 12th grade -
**bobpursley**, Tuesday, January 6, 2009 at 9:36pmnot basically 20, it is 20. Units were not specified.

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:36pmoh okay Thank You

- 12th grade -
**bobpursley**, Tuesday, January 6, 2009 at 9:37pmWHOA> STOP.

the v(t) is only valid for t Less than t=0. So the answer is unable to determine.

- 12th grade -
**bobpursley**, Tuesday, January 6, 2009 at 9:38pmthe question makes no sense. It starts at time zero, and there is no definition of velocity after that.

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:40pmso t less than or equal to 0 is invaild

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:42pmI copied the problem out of my textbook

- 12th grade -
**bobpursley**, Tuesday, January 6, 2009 at 9:42pmThe problem asks for postion at time 4, which is outside the time given for the velocity expression, if you typed it correcty. One needs the velocity function for time zero to somewhat beyond 4 to calculate postion.

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:44pmthe velocity v(t)= 2t + 1

- 12th grade -
**bobpursley**, Tuesday, January 6, 2009 at 9:46pmbarbara, you originally asked for position at time 4 for a function that has a certain velocity function before time zero. One cant calculate postition for time 4 unless there is some indication of what it was doing between time zero and time 4.

- 12th grade -
**Barbara**, Tuesday, January 6, 2009 at 9:49pmoh okay well Thank You for you help I will ask my teacher how to solve this tomorrow

- 12th grade -
**drwls**, Wednesday, January 7, 2009 at 1:43amI think you misstated the problem. Is it really

"A particle starts at x=0 and moves along the x-axis with velocity v(t)= 2t+1 for time t GREATER than or equal to 0. Where is the particle at t=4?" ?

If so, position is the integral of velocity from t=0 to t = 4.

That integral is the change in t^2 + t from t = 0 to 4, which is 20.

You can get the same answer by multiplying the average velocity for the interval, which is 5, but the time interval (4).