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# Analyzing motion problems: position

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.C (LO)
,
CHA‑4.C.1 (EK)

## Video transcript

- [Instructor] Divya received the following problem. A particle moves in a straight line with velocity v of t is equal to the square root of three t minus one meters per second, where t is time in seconds. At t equals two, the particle's distance from the starting point was eight meters in the positive direction. What is the particle's position at t equals seven seconds? Which expression should Divya use to solve the problem? So pause this video and have a go at it. Alright now let's do this together. So we wanna know the particle's position at t is equal to seven. So what we could do, they tell us what our position is at t equals two, so what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven. And there's another word for this. You could also call this your displacement from t equals two to t equals seven. And we know how to think about displacement. Velocity is your rate of change of displacement. And so if you wanna figure out your displacement, your between two times, you would integrate the velocity function. So this is going to be the integral from t equals two to t equals seven of our velocity function, v of t, dt. This would be our displacement from time two to time seven. So if they said what is our change in position from time two to time seven, it would be just this expression. But they want us, they want us to figure out, or they want Divya to figure out, what is the particle's position at t equals seven seconds. So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is. It was eight meters in the positive direction, so we could just call that positive eight meters. So it's going to be eight plus your change in position, which is going to be your displacement. And let's see we can see this choice right over there, and that's what we would pick. This first option, v of seven, that just gives us our velocity at time seven, or at the exactly at seven seconds. Or another way, our rate of change of displacement at seven seconds. So that's not what we want. This one right over here, you have your position at t equals two, but then you have your change in position from t equals zero to t equals seven. So this doesn't seem, this isn't right. And this is your position at time two plus your v prime, the derivative of velocity is the acceleration, plus your acceleration at time seven, so that's definitely not gonna give you your the particle's position. So we like that second choice.
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