Posted by Barbara on .
A particle starts at x=0 and moves along the xaxis with velocity v(t)=2t+1 for time t is less than or equal to 0. Where is the particle at t=4?

12th grade Subject?? 
Ms. Sue,
What is your subject?

12th grade 
Barbara,
Calculus

12th grade 
bobpursley,
I 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,
so my answer would basically be 20

12th grade 
bobpursley,
not basically 20, it is 20. Units were not specified.

12th grade 
Barbara,
oh okay Thank You

12th grade 
bobpursley,
WHOA> STOP.
the v(t) is only valid for t Less than t=0. So the answer is unable to determine. 
12th grade 
bobpursley,
the question makes no sense. It starts at time zero, and there is no definition of velocity after that.

12th grade 
Barbara,
so t less than or equal to 0 is invaild

12th grade 
Barbara,
I copied the problem out of my textbook

12th grade 
bobpursley,
The 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,
the velocity v(t)= 2t + 1

12th grade 
bobpursley,
barbara, 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,
oh okay well Thank You for you help I will ask my teacher how to solve this tomorrow

12th grade 
drwls,
I think you misstated the problem. Is it really
"A particle starts at x=0 and moves along the xaxis 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).