Velocity is a vector, meaning it has both magnitude and direction. The "direction" of velocity is either positive or negative. Positive and negative velocities describe motion in opposite directions. In Dance Dance Revolution, negative would either be juking left or back and positive would be moving towards the forward and right arrows. However, if the player would rotate 90 degrees, the relative negative and positive directions might change. DDR would enter bazaar-o world, but it could happen. The context of a problem will tell you how to interpret positive and negative velocities for that problem.
Sample Problems
1) Let v(t) be the velocity of a bug crawling back and forth on the x-axis. When v(t) is positive, the bug is moving to the right:
When v(t) is negative, the bug is moving to the left.
2) Calvin is biking away from his house with velocity v(t).
When v(t) is negative it means Calvin is biking back towards his house.
3) A bird is flying North with velocity v(t).
When v(t) is negative it means the bird is flying South.
Speed is the magnitude, or absolute value, of velocity. This means speed can't be negative!
Sample Problem
Let v(t), in units per second, be the velocity of a bug crawling back and forth on the x-axis. When v(t) = -5 it means the bug is crawling left at a speed of 5 units per second.
When a question asks "how fast" something is going, you're being asked for the speed.
Sample Problem
When Calvin's velocity is -7 miles per hour, how fast is he going?
Answer.
The question is asking for Calvin's speed, which is 7 miles per hour. The negative sign tells which direction Calvin is traveling, but doesn't have anything to do with his speed.
Now that we know what negative velocities mean, it's time to bring the integrals into it again and figure out what integrals of negative velocities mean.
When velocity is non-negative, we know that
When velocity is negative, the integral of velocity is also negative. We can think of this negative value as a weighted distance.
The number part tells the distance, and the negative sign tells the direction, which is the opposite of the direction travelled when v(t) is positive.
Take the area between v(t) and the t-axis that's above the axis. This is the total distance the bug travels to the right.
Take the area that's between v(t) and the t-axis, below the t-axis. This is the total distance the bug travels to the left.
Subtract:
(distance bug travels right) – (distance bug travels left) = 9 – 19 = -10.
This is the net change in the bug's position from time t = 0 to time t = 10.
We found this net change by taking a weighted sum of the areas between v(t) and the t-axis, which means we also just found the integral of v(t) from 0 to 10:
Whatever the units, when velocity is 0, speed is also 0. If something isn't moving at all, it's not moving in any direction. If v(t) is Calvin's velocity away from his house, measured in mph, then when v(t) = 0 Calvin isn't moving away from his house and he's not moving towards his house either.
Practice:
Brooke is strolling through the forest. Her velocity East is given by the following graph: 
(a) At what time(s) is Brooke not moving? (b) On what time interval(s) is Brooke moving East? (c) On what time interval(s) is Brooke moving West? (d) At what time(s) is Brooke moving most rapidly East? (e) At what time(s) is Brooke moving most rapidly West? (f) At what time(s) is Brooke moving most rapidly? (g) On what time interval(s) is Brooke speeding up? (h) On what time interval(s) is Brooke slowing down? | |
When v(t) is positive, Brooke is moving East. When v(t) is negative, Brooke is moving West. When v(t) = 0, Brooke isn't moving at all. (a) Whenever v(t) = 0, Brooke is not moving. This occurs when t = 0, 3, 6, and 10. (b) Brooke is moving East whenever v(t) > 0. This corresponds to the intervals 0 < t < 3 and 6 < t < 10. We use the < sign instead of the ≤ sign because when t equals 0, 3, 6, or 10, Brooke isn't moving at all. Thus we we don't want to include those values of t in the intervals. (c) Brooke is moving West whenever v(t) < 0. This corresponds to the interval 3 < t < 6. (d) Brooke is moving most rapidly East when v(t) > 0 and |v(t)| is as large as possible (since speed is the absolute value of velocity). The largest positive value of velocity is v(2) = 3, so Brooke is moving most rapidly East when t = 2. (e) Brooke is moving most rapidly West when v(t) < 0 and |v(t)| is as large as possible. This means we're looking for the most negative value of velocity, which occurs when t = 4. (f) Brooke is moving most rapidly at t = 4, since |v(4)| = 4 mph is the largest absolute value the velocity attains on the time interval [0,10]. (g) Since speed is the absolute value of velocity, Brooke is speeding up whenever her velocity is getting farther away from zero. This occurs on the intervals (0,2), (3,4), and (6,9). (h) Brooke is slowing down whenever her velocity is getting closer to zero, since this means her speed |v(t)| is also getting closer to zero. This occurs on the intervals (2,3), (4,6), and (9,10). | |
A cat starts at ground level and climbs a tree with velocity given by the graph below. When v(t) > 0 the cat is climbing upwards. 
How high is the cat when t = 4? When t = 7? | |
From t = 0 to t = 4 the area between v(t) and the t-axis is 8. Since v(t) is non-negative on this interval, the distance the cat travels from t = 0 to t = 4 is 
This means when t = 4 the cat is 8 feet above the ground. From t = 4 to t = 7 the area between the graph of v(t) and the t-axis is 4.5. Since v(t) is negative on this interval (except at the endpoints), the integral will also be negative. The cat travels a weighted distance of 
Because of the negative sign, the cat is now traveling the opposite direction, so from t = 4 to t = 7 the cat travels 4.5 feet downwards. The cat ends up at 8 – 4.5 = 3.5 feet above the ground. | |
A squirrel starts two feet above ground and climbs a tree with velocity given by the graph below. When v(t) > 0 the squirrel is climbing upwards. 
How high is the squirrel when t = 7? | |
This is the same graph from the previous example, so the values of the integrals will be the same. From t = 0 to t = 4 the squirrel climbs upward  .
From t = 4 to t = 7 the squirrel climbs a weighted distance of , meaning it climbs down 4.5 feet. Since the squirrel started at a height of 2 feet, when t = 7 the squirrel's height above ground is 2 + 8 – 4.5 = 5.5 feet. | |
A bug is crawling around on a number line. The bug's velocity, in units per second, is given by the graph below. Positive velocities indicate that the bug is traveling to the right. 
Assume the bug is at 0 on the number line when t = 0 seconds. (a) Where on the number line is the bug when t = 4,6,8, and 10? (b) At what time(s) does the bug change direction? (c) How many times during the interval 0 ≤ t ≤ 10 is the bug at position -8 on the number line? | |
(a) We can see from the graph that 
This means from time t = 0 to time t = 4, the bug travels 13 units to the left, so at t = 4 the bug is at -13 on the numberline. From t = 4 to t = 6 the bug travels 
to the right, so at t = 6 the bug is at position -13 + 4 = -9. From t = 6 to t = 8 the bug travels 
to the right, so at t = 8 the bug is at -9 + 5 = -4. From t = 8 to t = 10 the bug travels a weighted distance of  ,
meaning the bug travels 6 units to the left. At t = 10 the bug is at -4 – 6 = -10. (b) Looking at the bug's journey, we can see that the bug changes direction at t = 4 and at t = 8. These are the times at which the graph of v(t) changes from negative to positive, and from positive to negative, respectively. (c) To see how many times during the interval 0 ≤ t ≤ 10 the bug is at -8, look at the bug's journey again. It passes -8 once on its way from 0 down to 13. It passes -8 a second time between t = 6 and t = 8, on its way from -9 up to -4. And it passes -8 a third time on its way down from -4 to -10. In all, the bug passes -8 three times during the interval 0 ≤ t ≤ 10. | |
Mara went on an hour-long bicycle ride. She started 15 miles from
home and her velocity for the hour is given by the graph below. 
When velocity is positive, it means her distance from home is increasing. How far from home was Mara after (a) 20 minutes? (b) 50 minutes? (c) one hour? | |
(a) 20 minutes is  of an hour. From t = 0 to  we can see from the graph that Mara travels 2.5 miles towards her home: Since she started 15 miles from her home, when  she is 15 – 2.5 = 12.5 miles from her home. (b) 50 minutes is  of an hour, so this question is asking where Mara is when  . From  to  she travels 5 miles away from her home. This means when  her distance from home is 12.5 + 5 = 17.5 miles. (c) From  to t = 1 Mara travels approximately .83 miles towards her home. This means after the hour, her distance from home is approximately 17.5 – 0.83 = 16.67 miles. | |
When v(t) = -100 feet per second, the cheetah is running South at a speed of 100 feet per second. When v(t) = 10, what direction is the cheetah running?
Answer
If negative velocity means the cheetah is running South, then positive velocity means the cheetah is running North.
If Jen's velocity is -60 mph, how fast is she going?
Answer
The absolute value of -60 is 60, so Jen is going 60 mph.
If a garden snail's velocity is 0 feet per second, how fast is the snail moving?
Answer
If the snail's velocity is 0 feet per second, the snail's speed is also 0 feet per second. This means the snail isn't moving at all.
Let v(t) be the velocity of a bug moving on the x-axis, measured in units per second. What does it mean to say v(t) = -4?
Answer
The bug is moving left at a speed of 4 units per second.
A hummingbird is flying back and forth with velocity given by the graph below, where positive velocities mean the hummingbird is flying North and negative velocities mean the hummingbird is flying South:
(a) At what time(s) is the hummingbird moving neither North nor South?
(b) On what time interval(s) is the hummingbird moving North?
(c) On what time interval(s) is the hummingbird moving South?
(d) At what time(s) is the hummingbird moving most rapidly North?
(e) At what time(s) is the hummingbird moving most rapidly South?
(f) At what time(s) is the hummingbird moving most rapidly?
(g) On what time interval(s) is the hummingbird speeding up?
(h) On what time interval(s) is the hummingbird slowing down?
Answer
(a) The hummingbird is moving neither North nor South whenever v(t) = 0, which occurs at t = 2, 5, and 7.
(b) The hummingbird is moving North when v(t) is positive, which occurs on the interval (2,5).
(c) The hummingbird is moving South when v(t) is negative, which occurs on the intervals [0,2) and (5,7). We do include t = 0 in the first interval, since v(0) is negative.
(d) The hummingbird is moving most rapidly North when v(t) is positive and as large as possible. This is when t = 4.
(e) The hummingbird is moving most rapidly South when v(t) is at its most negative. This occurs at both t = 0 and t = 6.
(f) The hummingbird is moving most rapidly when |v(t)| is largest. This occurs when t = 4.
(g) The hummingbird is speeding up on all intervals where v(t) is getting farther from zero (this means its speed |v(t)| is getting bigger). For this graph, v(t) is getting farther from zero on the intervals (2,4) and (5,6).
(h) The hummingbird is slowing down on all intervals where its speed |v(t)| is getting smaller, meaning v(t) is getting closer to zero. In this case, v(t) is getting closer to zero on the intervals (0,2), (4,5), and (6,7).
A bug crawling on a number line has velocity given by the graph below, where positive velocities indicate the bug is crawling to the right:
(a) At what time(s) does the bug change direction?
(b) On what time interval(s) is the bug moving to the right?
(c) On what time interval(s) is the bug moving to the left?
(d) At what time(s) is the bug moving most rapidly to the right?
(e) At what time(s) is the bug moving most rapidly to the left?
(f) At what time(s) is the bug moving most rapidly?
(g) On what time interval(s) is the bug speeding up?
(h) On what time interval(s) is the bug slowing down?
Answer
(a) The bug changes direction whenever v(t) goes from positive to negative, which happens at t = 4 and t = 12, or from negative to positive, which happens at t = 8. We do NOT include t = 0, because we can't tell if v(t) is negative to the left of t = 0. To summarize, the bug changes direction at t = 4, 8, and 12.
(b) The bug is moving to the right whenever v(t) is positive, which occurs on the intervals (0,4) and (8,12).
(c) The bug is moving to the left whenever v(t) is negative, which occurs on the intervals (4,8) and (12,16).
(d) The bug is moving most rapidly to the right when t = 10, since this is where the largest positive value of v(t) occurs.
(e) The bug is moving most rapidly to the left when t = 16, since this is where the most negative value of v(t) occurs.
(f) The bug is moving most rapidly when t = 16, since this is where |v(t)| is largest (equivalently, this is where v(t) is farthest from 0).
(g) The bug is speeding up on (0,2), (4,6), (8,10), and (12,16), since these are the intervals where v(t) is getting further from 0.
(h) The bug is slowing down on (2,4), (6,8), and (10,12), since these are the intervals where v(t) is getting closer to 0.
A cat climbs up and down a tree with velocity given by the graph below. When the velocity is positive it means the cat is climbing up the tree.
(a) At what time(s) does the cat change direction?
(b) At what time(s) is the cat climbing most rapidly upwards?
(c) At what time(s) is the cat climbing most rapidly downwards?
(d) What is the cat's fastest speed on the interval [0,7]?
(e) What is the cat's slowest speed on the interval [0,7]?
(f) At what time(s) is the cat highest in the tree?
Answer
(a) The cat changes direction whenever its velocity changes from positive to negative or from negative to positive. The only time this happens on [0,7] is at t = 5, where the cat changes from climbing up the tree (positive velocity) to climbing down the tree (negative velocity).
(b) The cat is climbing most rapidly upwards when t = 0 and when t = 4, since these are the times when the cat's velocity is largest.
(c) The cat is climbing most rapidly downwards when t = 6, since this is when the cat's velocity is most negative.
(d) The cat's fastest speed on the interval [0,7] is 2 feet per second, which occurs at t = 0 and t = 4.
(e) The slowest the cat ever goes on [0,7] is 0 feet per second.
(f) From t = 0 to t = 5 the cat is climbing up, and from t = 5 to t = 7 the cat is climbing down. The cat is highest in the tree when t = 5, since this is when it stops climbing up and starts climbing down.
A bug crawls back and forth on a number line with velocity given by the graph below. Positive velocities correspond to movement in the positive direction on the number line.
(a) If the bug is at 0 on the number line when t = 0, determine the position of the bug at t = 3, 5, and 8 seconds.
(b) If the bug is at -10 on the number line when t = 0, determine the position of the bug at t = 3, 5, and 8 seconds.
Answer
From t = 0 to t = 3 the bug travels
to the right.
From t = 3 to t = 5 the bug travels a weighted distance of
,
meaning it travels 3 units to the left.
From t = 5 to t = 8 the bug travels
to the right.
(a) At t = 0 the bug is at 0 on the numberline.
From t = 0 to t = 3 the bug moves 5 units right, so at t = 3 the bug is at 5.
From t = 3 to t = 5 the bug moves 3 units left, so at t = 5 the bug is at 2.
From t = 5 to t = 8 the bug moves 10 units right, so the bug ends up at 12
on the numberline when t = 8.
(b) The bug will be moving the same distances, but with a different starting place.
At time t = 0 the bug is at -10. From t = 0 to t = 3 the bug moves right 5, so when t = 3 the bug is at
-10 + 5 = -5.
From t = 3 to t = 5 the bug moves left 3, so when t = 5 the bug is at
-5 – 3 = -8.
From t = 5 to t = -8 the bug moves right 10, so when t = 8 the bug is at
-8 + 10 = 2.
The graph below describes the velocity of a car over a 10-hour scenic drive. Positive velocity indicates the car is traveling East.
(a) Select the correct answer: From time t = 0 hours to t = 4 hours the car is traveling (East|West).
(b) Fill in the blanks: From time t = ? to time t = ? the car is traveling West.
(c) At what time(s) is the car stopped?
(d) After 10 hours of driving, is the car East or West of where it started? How far East or West of where it started?
Answer
(a) From time t = 0 to t = 4 the car is traveling East because the graph of v(t) is positive on (0,4).
(b) From time t = 6 hours to time t = 10 hours the car is traveling West, because the graph of v(t) is negative on (6,10).
(c) The car is stopped whenever its velocity is 0, which occurs at t = 4, 6, and 10 hours.
(d) By looking at the graph of v(t), we can see that from t = 0 to t = 6 the car travelled a total of 120 miles East:
From t = 6 to t = 10 the car travelled 80 miles West:
The car's net change in position was
120 – 80 = 40
miles to the East, so the car ended up 40 miles East from where it started.
A whiny toddler is in the exact center of a 20-foot long room. At one end of the room is a lollipop and at the other end is a teddy bear. The toddler toddles back and forth with velocity given by the graph below. When velocity is positive, it means the toddler is moving towards the teddy bear.
(a) Describe the toddler's location at t = 2, 4, 6, and 10 seconds.
(b) How many times between t = 0 and t = 10 seconds will the toddler pass through the exact center of the room?
(c) Will the toddler eventually reach the lollipop, the teddy bear, or neither?
Answer
First we need to understand the set-up. A toddler is between a lollipop and a teddy bear, each of which is 10 feet away from the toddler. We put the teddy bear on the right side because the problem says positive velocities mean the toddler is moving towards the teddy bear. We're used to thinking of positive velocities as moving us to the right on a number line, so if we put the teddy bear on the right side of the room, positive velocity still corresponds to moving right.
(a) We look at the graph to see how far the toddler moves over each time interval.
From t = 0 to t = 2 seconds the toddler moves
to the right, so the toddler is 14 feet from the lollipop and 6 feet from the teddy bear:
From t = 2 to t = 4 seconds the toddler moves
which means it moves 2 feet to the left (closer to the lollipop). The toddler is now 12 feet from the lollipop and 8 feet from the teddy bear.
From t = 4 to t = 6 seconds the toddler moves
closer to the bear, so the toddler is now 15 feet from the lollipop and 5 feet from the bear.
From t = 6 to t = 10 the toddler moves
meaning 6 feet to the left (closer to the lollipop). The toddler is now 9 feet from the lollipop and 11 feet from the bear.
What we did here was put the lollipop, toddler, and bear on a number line. The lollipop was at -10, the toddler at 0, and the bear at 10.
There are other ways to do this problem. For example,
we could put the lollipop at 0, the bear at 20, and start the toddler off at 10.
Then the toddler would end up at
10 + 4 – 2 + 3 – 6 = 9
on the numberline, meaning 9 feet from the lollipop and 11 feet from the bear.
Another possibility is that we could put the lollipop at -20, the bear at 0, and the toddler at -10:
Any way we do it, the toddler will end up 9 feet from the lollipop and 11 feet from the bear.
(b) The toddler is at the center of the room when t = 0, and passes through the center of the room once more between t = 6 and t = 10, in order to get closer to the lollipop than to the bear. This means the toddler passes through the center of the room twice.
(c) The toddler never reaches the lollipop or the bear. The toddler gets within 5 feet of the bear, and within 9 feet of the lollipop, but never gets to either.
