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

Calculus

I assume you are in calculus.

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

Put in t=4 and compute.

so my answer would basically be 20

not basically 20, it is 20. Units were not specified.

oh okay Thank You

WHOA> STOP.

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

the question makes no sense. It starts at time zero, and there is no definition of velocity after that.

so t less than or equal to 0 is invaild

I copied the problem out of my textbook

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.

the velocity v(t)= 2t + 1

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.

oh okay well Thank You for you help I will ask my teacher how to solve this tomorrow

I 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).

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