Question
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?
Answers
Ms. Sue
What is your subject?
Barbara
Calculus
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.
Postition=INTEGRAL v(t) dt=INT (2t+1)dt
= t^2 + t
Put in t=4 and compute.
Barbara
so my answer would basically be 20
bobpursley
not basically 20, it is 20. Units were not specified.
Barbara
oh okay Thank You
bobpursley
WHOA> STOP.
the v(t) is only valid for t Less than t=0. So the answer is unable to determine.
the v(t) is only valid for t Less than t=0. So the answer is unable to determine.
bobpursley
the question makes no sense. It starts at time zero, and there is no definition of velocity after that.
Barbara
so t less than or equal to 0 is invaild
Barbara
I copied the problem out of my textbook
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.
Barbara
the velocity v(t)= 2t + 1
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.
Barbara
oh okay well Thank You for you help I will ask my teacher how to solve this tomorrow
drwls
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).
"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).