In addition to finding functions that model an object’s movement, you may be asked to analyze its motion given a position, velocity, or acceleration graph. Here are key points to consider when reading graphs:
Position graph:
The slope of the tangent line at any point on the position graph represents the instantaneous velocity.
An increasing graph means velocity is positive, and the object is moving right/forward/up.
A decreasing graph means the velocity is negative.
Horizontal tangents or lines (zero slope) mean the object is at rest.
Velocity graph:
The slope of the tangent line at any point on the velocity graph represents the instantaneous acceleration.
Check whether the graph is above or below the x-axis to determine the velocity’s sign.
Combine this with the sign of the acceleration to determine if the object is speeding up or slowing down over a period of time.
1. A particle moves along the number line with its position p(t) plotted as a function of time (in seconds).
Graph of p(t)
a) What is its average velocity from t=1 to t=3 seconds?
b) When is the particle at rest?
c) When is it moving left and when is it moving right?
Solutions
a) What is its average velocity from t=1 to t=3 seconds?
The average velocity is the change in position over the change in time.
vavg=3−1p(3)−p(1)
From the graph, p(3)=−2. We can find the exact value of $p(1): the straight line in between points (0,0) and (3,−2) is defined by the equation p(t)=−32t.
So p(1)=−32 and the average velocity is
vavg=3−1−2−(−32)
=2−34
=−32
The average velocity over [1,3] is −32 units/sec. It’s negative since the particle’s position in the first 3 seconds is negative, meaning it moved to the left.
b) When is the particle at rest?
A particle is at rest when its velocity is zero. On the position graph, we’re looking for the time interval where the slope is zero - horizontal lines. These occur over the intervals (3,5) and (7,9).
c) When is it moving left and when is it moving right?
The particle is moving left when v(t)<0 and right when v(t)>0.
We have to look at the slopes of the tangent lines to the position graph. Here’s a sign chart for v(t) with each interval representing a piece of the position graph, if it were written as a piecewise function:
Interval
Sign of v(t)
(0,3)
−
(3,5)
0
(5,7)
+
(7,9)
0
(9,11)
+
From the sign chart, the particle’s movement is as follows:
It moves left on 0<t<3
Stops on 3<t<5
Moves right on 5<t<7
Stops again on 7<t<9
Continues moving right on 9<t<11
2. The velocity of a particle moving along a number line, in meters per minute, is displayed in the graph below.
Graph of v(t)
a) At what time(s) does the particle change direction?
b) When is it moving at constant speed?
c) When is it speeding up and when is it slowing down?
d) What is the particle’s maximum speed, and when does this occur?
Solutions
a) At what time(s) does the particle change direction?
(spoiler)
The particle changes direction when the sign of the velocity changes from positive to negative or vice versa.
This time, we look for whether the velocity is above or below the x-axis. When the function crosses the x-axis, it will have an x-intercept.
The first x-intercept is (3,0).
The second is unclear but we know the piece goes through (7,−3) and (10,−2). Finding the equation of a line between two points gives v(t)=35(x−10)+2. The x-intercept occurs where y (or v(t)) is 0. Solving, it’s at t=8.8 minutes.
The third x-intercept is at (12,0).
The organized sign chart shown below separates intervals based on the x-intercepts.
Interval
Sign of v(t)
(0,3)
+
(3,8.8)
−
(8.8,12)
+
(12,14)
+
v(t) changes sign (which means the particle changes direction) at t=3 and t=8.8 minutes only.
b) When is it moving at constant speed?
(spoiler)
A particle is moving at constant speed when its acceleration is zero while the velocity is not zero. Since acceleration is the derivative of velocity, we’re looking for portions where the slope of the velocity graph is zero. This occurs in the interval (4,7). In that interval of time, the velocity is −3 meters/minute, meaning it’s moving to the left at a speed of 3 meters per minute.
c) When is it speeding up and when is it slowing down?
(spoiler)
A particle speeds up when its acceleration and velocity have the same sign and slowing down when the signs oppose.
We look at whether the velocity graph is above or below the x-axis to determine the sign of the velocity, and whether the slope of the line segment is positive, negative, or zero to determine the sign of the acceleration.
Here’s a sign chart for with intervals separated according to line segments and x-intercepts.
Interval
Sign of v(t)
Sign of a(t)
(0,2)
+
+
(2,3)
+
−
(3,4)
−
−
(4,7)
−
0
(7,8.8)
−
+
(8.8,10)
+
+
(10,12)
+
−
(12,14)
+
+
The particle is
Speeding up in:
(0,2)
(3,4)
(8.8,10)
(12,14)
Slowing down in:
(2,3)
(7,8.8)
(10,12)
d) What is the particle’s maximum speed, and when does this occur?
(spoiler)
The particle hits a maximum speed of 3 meters/minute. However, this happens at several points - once at t=2 minutes, another time between 4 and 7 minutes (traveling at constant speed to the left instead) and again at t=14 minutes.
3. A particle’s position along the x-axis is tracked at specific points in time and recorded in the following table:
Time t (seconds)
Position (meters)
0
5
1
13
2
18
3
20
4
16
5
10
a) What is the average velocity from t=2 to t=5 seconds?
b) Estimate when the particle changes direction.
a) Average velocity over [2,5]
Solution
(spoiler)
vavg=5−2p(5)−p(2)
=310−18
=−38 m/s
b) Estimate when the particle changes direction.
Solution
(spoiler)
Looking at the table numbers:
From t=0 to t=3, the position increases (5→20), meaning the particle moves forward.
From t=3 to t=4, the position decreases (20→16), meaning the particle moves backward.
So the particle must have changed direction at about 3 seconds.
4. A particle’s moves along the x-axis with its velocity tracked at specific points in time and recorded in the following table:
Time t
Velocity (m/s)
0
0
1
5
2
7
3
2
4
−3
5
−9
a) Estimate the rate of change of velocity at t=2 seconds.
b) Estimate when the particle changes direction.
a) Rate of change of velocity at t=2 seconds
Solution
(spoiler)
While the velocity at t=2 is 7 m/s, the rate of change of velocity is about estimating the acceleration. We can use the two surrounding points for a good estimate.
aavg=3−1v(3)−v(1)
=22−5
=−23 m/s2
b) Estimate when the particle changes direction.
Solution
(spoiler)
A particle changes direction when its velocity changes sign, meaning. it will be 0 at some point. Since the velocity at t=3 seconds is positive and at t=4 seconds it’s negative, then by the Intermediate value theorem there must be a point on (3,4) where the velocity is 0.
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