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a particle moving moves along a number line not shown 40 is greater than or equal to zero its position function s of T is shown in blue so this is its position as a function of time its velocity function V of T is in red its velocity and its acceleration function a of T is in green all are graphed with respect to time T in seconds with the graphs as an aid answer the questions below so that's what's going on here so its position as a function of time actually let me just draw their number line that they did not depicts just so we can really think about this so let's say this is our number line let's say that this right over here is the that's 0 that's 1 that is 2 this is negative 1 so we're defining going to the right as the positive direction so what's happening here so at time equals 0 at time equals 0 right over here time equals 0 s of 0 is 0 and then as time increases our position increases all the way until time equals 1 at time equals 1 our position is 2 so time equals 1 our position is 2 and then our position and then s of T starts decreasing so we start it one way to think about it is and you see we move up we move we move to the right really fast we get to 2 we stop at 2 and then we start moving to the left so one thing so time equals 0 the first second looks like this resume well slow down and stop and then we start moving the other way and then we start drifting notice our position is good is is decreasing so our position is decreasing but it's decreasing it ever slower slower and slower and slower rates it's not clear if we'll ever get back to the origin so that's what's going on here and we see that no matter which graph we look at our position function is definitely telling that story our velocity function which is the derivative of the position function is telling you that story out the gate we have a we have a high positive velocity but we decelerate quickly and at 1 second our velocity is 0 and then we start having a negative velocity which means we're moving to the left so fast rightward velocity but we decelerate faster driver to decelerate quickly stop at time equals one second and then we start drifting to the left and the acceleration also also shows that same narrative but anyway let's actually answer the questions the initial velocity of the particle is blank units per second I encourage you to pause this video and answer that well we just said the velocity lets see a time equals zero we're at eight units per second so I'll just put eight right over there the particle is moving to the right when T is a part of or when T is in the interval and they're since they're doing this as a member of they really want this kind of in the set notation T is a member of the interval well when are we moving to the right we already went over that we're moving to the right there's a couple of ways to think about it when our velocity is greater than so we're moving to the right when V of T is greater than zero when V of T is less than zero we're moving to the left when V of T is equal to zero with stationary so when is V of T greater than zero well it's between T is between T being zero velocity is definitely positive all the way to T is one but not including T is one so I'll put a parenthesis there so this is equivalent to saying so T is a member of that interval is equivalent to saying that zero is less than or equal to T is less than is less than one once again the first second we're going at time zero we're going fast slowed down and then stopped for a infinitesimal moment and then we start drifting back that happens at time equals one we start drifting back the total distance traveled by the particle for T in the interval between zero and three is blank units so once again I encourage you to pause the video and try to answer that the total distance so this is interesting don't get consistence confused with displacement if I if I were to move three to the right and then I were to move back one to the left the total distance I've traveled is four the distance I traveled is four while the displacement while the displacement would be a positive and we could make put a minus 1 there we moved one to the left so three to the right one to the left while our displacement would be a net of positive to our displacement would be a positive two so they're asking what's the total distance traveled so between time 0 and time 1 between time 0 and time 1 we have moved we have moved to to the right we've moved to to the right we've moved to to the right and then between time 1 and time 3 we move back or to the left we move half so to the left we move we move half so what's our total distance it's going to be 2 to the right plus 1/2 to the left which is going to be 2 point 5 units 2 to the right plus 1/2 to the left 2 point 5 units and we're done