Main content

### Course: Calculus 2 > Unit 4

Lesson 2: Straight-line motion- Motion problems with integrals: displacement vs. distance
- Analyzing motion problems: position
- Analyzing motion problems: total distance traveled
- Motion problems (with definite integrals)
- Analyzing motion problems (integral calculus)
- Worked example: motion problems (with definite integrals)
- Motion problems (with integrals)
- Average acceleration over interval

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Analyzing motion problems: position

Finding the appropriate expression to use when looking for the position at a certain time.

## Want to join the conversation?

- We can't calculate the definite integral of this function from 0 to 7, could we ? as it's undefined for t less than 1/3(8 votes)
- Yes that cannot be done, that is why they do the definite integral from 2 to 7 in the video.(8 votes)

- the velocity of the particle v(t) = sqrt(3t-1)

so, v(t) =sqrt(3) * sqrt(t - 1/3 ) so depends on the equation the velcoty of the partcle starts at t = 1/3 with velocty = sqrt(3)

the net change of position from t =1/3 to t =2 is equal ((10*sqrt(5))/9)

So the question of how the distance from the starting point to t= 2 is equal to 8?(2 votes)- The velocity of the particle doesn't follow the model at t<2(2 votes)

- the particle's position is undefined when t is less than 1/3. What would it look like at that time? Would it just not move ?(2 votes)
- We don't know. The particles motion before t=1/3 is beyond the scope of this model, so it can't tell us anything about that time period.(2 votes)

- If the equation of the velocity holds, what possibly happened in the 1st second? Why not use a piece-wise equation instead?(1 vote)
- Well, you could just integrate the velocity function, which will give you a position function. Using that, you can find the position at t=1 (which would also be its displacement). Note that you have the value of displacement at t=2, so you can find the exact position function rather than one with the +C in it.

A piecewise function isn't needed here as, well, the velocity could be modelled by a single equation. If the velocity showed different trends over different intervals,*then*we would need to use piecewise functions to model it.(1 vote)

- Couldn't you just use the rate of change in position from time 0 to time 7 to find the displacement? So basically option C but without adding the time at t=2. I know that this isn't one of the options for answers for this question but if you were given the problem without multiple-choice answers couldn't you solve it this way?(1 vote)
- The function given to us has t<(1/3) undefined, so we
**can't**find the rate of change from 0 to 7. We do, however, already know the displacement after 2 seconds, so we would use that combined with the rate from 2 to 7 as in the video.(1 vote)

- Is the +8 similar to the +C term in indefinite integrals?(1 vote)
- Whenever he says position, do i just find the displacement?(1 vote)
- yes when he x-displacement he means x-coordinate, y-displacement he means y-coordinate(1 vote)

- if i integrate the velocity i would get the position function like this

Position = F(t) + c

if i know position at t=2 , couldn't i find C, and then plug any t to solve for position?(1 vote)- Correct. If you were told to solve it, that's what you'd do. Here though, Sal focuses more on the setup, which is really the more important part of calculus, as integration is easy enough.(1 vote)

## 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.