Negative Velocity Examples

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 6.

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 6 feet downwards.

The cat ends up at

8 – 6 = 2 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 6 feet. Since the squirrel started at a height of 2 feet, when t = 7 the squirrel's height above ground is

2 + 8 – 6 = 4 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 number line.

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 0.83 miles towards her home.

This means after the hour, her distance from home is approximately

17.5 – 0.83 = 16.67 miles.