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## Straight-line motion

# Analyzing straight-line motion graphically

## Video transcript

A particle moves along a
number line not shown for t is greater than or equal to 0. 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. That's velocity. And its acceleration
function, a if 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 it's position as
a function of time. Actually, let me just
draw their number line that they did not depict just so
we can really think about this. So let's say that's
our number line. Let's say this right
over here is 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 right
over here, 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 at time equals 1,
our position is 2. And then our position, s
of t, starts decreasing. So one way to think about it
is, and you see we move up, we move to the
right really fast. We get to 2. We stop at 2, and then we
start moving to the left. So at time equals 0, the
first second looks like this. We go zoom, oh,
slow down and stop. And then we start
moving the other way. And then we start drifting. Notice our position
is decreasing. So our position is
decreasing, but it's decreasing at 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 that story. Out the gate, 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 quickly, stop at time equals
1 second, and then we start drifting to the left. And the acceleration 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,
let's see, at time equals 0, we're at 8 units per second. So we'll just put
8 right over there. The particles moving
to the right when t is in the interval, and
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 it 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 0. When v of t is less than 0,
we're moving to the left. When v of t is equal
to 0, we're stationary. So when is v of
t greater than 0? Well, it's between t being
0, velocity is definitely positive, all the way to t is
1, but not including t is 1. So I'll put a parentheses there. So this is equivalent
to saying-- so t is a member
of that interval is equivalent to saying that
0 is less than or equal to t is less than 1. Once again, the first
second, at time 0, we're going fast, slow
down, and then stop for an infinitesimal moment,
and then we start drifting back. That happens at time equals 1. We start drifting back. The total distance traveled
by the particle for t in the interval between
0 and 3 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 distance
confused with displacement. 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 would be a positive. We could maybe put
a minus 1 there. We moved one to the left. So three to the right,
one to the left. Well, our displacement would
be a net of positive 2. So they're asking, what's
the total distance traveled? So between time 0 and time 1,
we have moved two 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 half. So what's our total distance? It's going to be
two to the right plus half to the left, which
is going to be 2.5 units. And we're done.