One Dimensional Motion
The Area Under the Curve
The Last Lecture
• During the last lecture, we discussed:
• The use of integration and algebra to derive the “BIG FIVE”
• Today we will discuss the significance of the “Area Under the Curve”
Let us recall the motion of the elevator.
The graph shows that the position vs. time
follows a non-linear (polynomial)
35
relationship
from 1-3 and 8-11 seconds,
and a constant change from 3-8 seconds.
y = -0.65t2 + 14.31t - 48.82
30
Position (m)
25
y = 4.00t - 8.00
20
With the exception of the first and last
second of the dataset, the motion of the
elevator can be broken up into three
distinct segments:
• Positive acceleration from 1 – 3 s
• Constant velocity from 3 – 8 s
• Negative acceleration from 8 – 11s
15
10
5
0
0
2
4
y = t2 – 2.00t + 1.00
6
8
Time (s)
10
12
14
If you calculate the area under the curve from 1 – 11 seconds, what do you get?
It should be clear by now that the area under a velocity vs. time curve equals the
displacement of the object.
4.5
4.0
Velocity (m/s)
𝑨𝟏 = ½ 𝒃 ∗ 𝒉
3.5
𝒅=½
𝒗∗𝒕
𝒅 = 𝒗𝒂𝒗𝒈 ∗ 𝒕
3.0
𝒅 = ½ (𝟒𝒎/𝒔)(𝟐𝒔)
𝑨𝟑 = ½ 𝒃 ∗ 𝒉
𝒅 = ½ (𝟒𝒎/𝒔)(𝟑𝒔)
𝒅 = 𝟔 𝒎𝒆𝒕𝒆𝒓𝒔
𝑨𝟐 = 𝒃 ∗ 𝒉
𝒅=𝒗∗𝒕
𝒅 = (𝟒𝒎/𝒔)(𝟓𝒔)
2.5
𝒅 = 𝟒 𝒎𝒆𝒕𝒆𝒓𝒔
2.0
𝒅 = 𝟐𝟎 𝒎𝒆𝒕𝒆𝒓𝒔
1.5
1.0
0.5
0.0
0
2
4
6
8
Time (s)
10
12
The total vertical displacement is equal to the sum of the areas: 𝑑𝑡𝑜𝑡𝑎𝑙 = 𝐴1 + 𝐴2 + 𝐴3 = 4𝑚 + 20𝑚 + 6𝑚 = 30𝑚
14
If you calculate the area under the curve from 1 – 3 and 8 – 11 seconds, what
do you get?
As should be clear once again, the area under an acceleration vs. time graphs equates to
the change in velocity of the object.
2.5
2.0
𝑨=𝒃∗𝒉
𝒗 = 𝒂 ∗ 𝒕
𝒗 = (𝟑𝒔)(−𝟏. 𝟑𝒎/𝒔𝟐)
𝒗 = ~ − 𝟒𝒎/𝒔
1.5
Acceleration(m/s2)
1.0
0.5
0.0
0
2
4
6
8
10
12
-0.5
𝑨=𝒃∗𝒉
𝒗 = 𝒂 ∗ 𝒕
-1.0
𝒗 = (𝟐𝒔)(𝟐𝒎/𝒔𝟐)
𝒗 = 𝟒𝒎/𝒔
-1.5
Note: If you-2.0
add the two areas together, you get zero, which tells Time
you(s)that the initial and final velocities should be the
same. In this case, it is zero.
14
• When the acceleration is uniform or constant, the area under a
velocity vs. time curve, and an acceleration vs. time curve are easy to
calculate.
• When the acceleration is zero or constant, the position of the particle will
follow a first or second order polynomial, respectively.
𝑥 − 𝑥𝑜 = 𝑣𝑜𝑡
Constant Velocity
𝑥 − 𝑥𝑜 = 𝑣𝑜𝑡 + 12𝑎𝑡2
2.5
00
22
44
66
302
Time (s)
(s)
Time
88
10
10
1
𝑣=2
𝑣 = 2.0 + 3.0𝑡
35
𝑥 = 2.0 + 2.0𝑡
𝑥 = 5.0 + 2.0𝑡 + 1.5𝑡2
Velocity
(m/s)
Velocity(m/s)
Position(m)
Position(m)
25
Second Order
25
1.5
20
𝐴 = 𝑏 × ℎ1
𝐴 = 𝑏×ℎ
2
151
10
0.5
5
0
0
𝐴=𝑏×ℎ
0
0
2
2
4
4
6
6
Time
Time (s)
(s)
𝑎=3
0.9
3.5
2)
(m/s2)
Acceleration
Acceleration(m/s
Constant Acceleration
200
180
20
160
140
15
120
100
10
80
60
5
40
20
00
First Order
8
8
10
10
0.8
3
0.7
2.5
0.6
2
0.5
𝑎=0
𝐴=𝑏×ℎ
0.4
1.5
0.3
1
0.2
0.5
0.1
0
0
0
0
22
44
66
Time
Time (s)
(s)
88
10
10
• However, what if the position vs. time of a particle does not follow a first or
second order polynomial relationship?
140
30
100
120
20
100
10
-6
0
-4
-2
-50
0
-100
-150
2
4
6
Velocity (m/s)
Position(m)
50
80
40
20
Acceleration not
constant
-4
-6
-4
-2
-20
-2
-10
0
-20
-30
0
-250
Time (s)
0
-6
60
-200
-300
Acceleration (m/s2)
150
-40
0
2
4
6
-50
Time (s)
Time (s)
𝑥 = 3𝑡 − 4𝑡2 + 𝑡3
Third Order
2
4
6
Now that we know what the area under the
curve represents, how do you calculate it?
We can break up the area into individual rectangles and add those areas up.
7
6
Velocity (m/s)
5
4
𝑨𝟐 = 𝒃 ∗ 𝒉
𝒙𝟐 = 𝒗 ∗ 𝒕
𝒙𝟐 = (𝟓. 𝟕𝟓𝒎/𝒔)(𝟏𝟎𝒔)
𝒙𝟐 = 𝟓𝟕. 𝟓𝒎
3
2
𝒙𝟑 = 𝟓𝟕. 𝟓𝒎
𝑨𝟏 = 𝒃 ∗ 𝒉
𝒙𝟏 = 𝒗 ∗ 𝒕
𝒙𝟏 = (𝟐. 𝟓𝒎/𝒔)(𝟏𝟎𝒔)
𝒙𝟏 = 𝟐𝟓𝒎
1
𝒙𝟏 = 𝟐𝟓𝒎
0
0
5
10
15
20
25
30
Time (s)
The total displacement is: 𝑥𝑡𝑜𝑡𝑎𝑙 = 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 25𝑚 + 57.5𝑚 + 57.5 + 25𝑚 = 165𝑚
35
40
Repeating the process using a smaller time interval
We will now reduce the time interval by a factor of 2 by going from 10s to 5s.
6
5
Velocity (m/s)
4
𝟓𝟗. 𝟕𝟓𝒎
𝟓𝟗. 𝟕𝟓𝒎
3
𝟓𝟐. 𝟎𝒎
2
𝟓𝟐. 𝟎𝒎
𝟑𝟔. 𝟕𝟓𝒎
𝟑𝟔. 𝟕𝟓𝒎
1
𝟏𝟑. 𝟓𝒎
𝟏𝟑. 𝟓𝒎
0
0
5
10
15
20
25
30
35
Time (s)
Recalculating the area: 𝑥𝑡𝑜𝑡𝑎𝑙 = 𝑥1 + 𝑥2 + ⋯ 𝑥8 = 13.5𝑚 + 36.75𝑚 + 52.0 + 59.75𝑚 + ⋯ = 162𝑚
40
We can repeat this process again and again infinitely producing a
smaller and smaller time interval each and every iteration.
7
6
Velocity (m/s)
5
4
3
2
1
0
0
5
10
15
20
Time (s)
25
30
35
40
Note that the area better approximates the displacement of the
object as the time interval decreases. This is especially noticeable
near the peak velocity.
7
𝑥 = lim
𝑡𝑛→0
6
𝑣𝑎𝑣𝑔 𝑡
𝑛
Velocity (m/s)
𝑣𝑎𝑣𝑔5
(for the interval)
4
3
2
1
0
0
5
10
𝑡𝑛
15
20
Time (s)
25
30
35
40
In calculus, if we know the formula, we can calculate
the area under the curve through integration
40
−0.0153𝑡2 + 0.6139𝑡 − 0.0697 𝑑𝑡
𝑥 =
𝑣 = −0.0153𝑡2 + 0.6139𝑡 − 0.0697
7
0
6
Velocity (m/s)
5
4
3
2
𝑥 = 0.0051𝑡3 + 0.3070𝑡2 − 0.0697𝑡
1
0
0
5
𝑥 =
10
−0.0051𝑚
𝑠3
15
40𝑠
3
+
20
25
Time (s)
0.3070𝑚
𝑠2
40𝑠
2
−
30
0.0697𝑚
𝑠
35
40𝑠 = 162𝑚
40