5.1 How do we measure distance travelled
# 5.1 How do we measure distance travelled?
A car moves along a straight road with increasing velocity.
| Time (sec) | Velocity (ft) |
|---|---|
| 0 | 20 |
| 2 | 30 |
| 4 | 38 |
| 6 | 44 |
| 8 | 48 |
| 10 | 50 |
| \(\Delta t = 2\text{ sec}\) |
\(\text{Distance} = \text{rate} * \text{time}\)¶
\(D = (60\text{ mph})(2\text{ hours})=120\text{ miles}\) For the first 2 seconds, we can underestimate the distance travelled using velocity at 0 seconds during 1st 2 seconds: \(d=(20)(2)=40\text{ ft}\) during next 2 seconds (from 2 to 4 seconds) \(d=(30)(2)=60\text{ ft}\)
Underestimate for distance travelled:¶
\((20)(2)+(30)(2)+(38)(2)+(44)(2)+(48)(2)\xcancel{+(50)(2)}\) \(=40+60+76+88+96\) \(=360\text{ ft}\)
Overestimate:¶
\(\xcancel{(20)(2)+}(30)(2)+(38)(2)+(44)(2)+(48)(2)+(50)(2)\) \(=60+76+88+96+100\) \(=420\text{ ft}\)
| Time (sec) | Velocity (ft) |
|---|---|
| 0 | 20 |
| 1 | 26 |
| 2 | 30 |
| 3 | 34 |
| 4 | 38 |
| 5 | 41 |
| 6 | 44 |
| 7 | 46 |
| 8 | 48 |
| 9 | 49 |
| 10 | 50 |
\(\Delta t = 1\text{ sec}\)
Underestimate¶
\(d=(20)(1)+(26)(1)+(30)(1)+(34)(1)+(38)(1)+(41)(1)+(44)(1)+(46)(1)+(48)(1)+(49)(1)\xcancel{+(50)(1)}\) \(=376\text{ ft}\)
Overestimate¶
\(d=\xcancel{(20)(1)+}(26)(1)+(30)(1)+(34)(1)+(38)(1)+(41)(1)+(44)(1)+(46)(1)+(48)(1)+(49)(1)+(50)(1)\) \(=406\text{ ft}\)
As the intervals get smaller, and the number of rectangles increase, the area of the rectangle approaches the area under the curve
Hokora's way of saying it: the smaller the iterations (\(\Delta t\)) is, the difference between the underestimation and overestimation approaches 0, and the area approaches the distance traveled
If velocity is positive, then total distance traveled is the area under the velocity curve!
Ex: With time \(t\) in seconds, the velocity of a bicycle is \(v(t)=5t\), in ft/sec. How far does the bike travel in first 4 seconds, after time \(t=0\)?¶
\(\text{Distance } = \text{ Area of triangle}\) \(=\frac{1}{2}(4)(20)\) \(=40\text{ ft}\)
Ex: Particle travels in straight line with velocity 30 cm/s for 5 seconds, and the -10 m/s for next 5 seconds (travels backwards). What do the following represent?¶
a) \(30*5+(-10)*5\)
Hokora's guess: distance from start
Written answer: \(\leftarrow\) change in position
b) \(30*5+10*5\)
Hokora's guess: distance traveled
Written answer: \(\leftarrow\) total distance traveled
If velocity can be negative as well as positive then: - change in position = area above axis - area below axis - total distance traveled = area above axis + area below axis
\(m=\frac{-150}{10}=-15\) \(v=-15t+100\) \(=\frac{20}{2}\)
Find
(a) change in position
\(\bigtriangleup 1 - \bigtriangleup 2\)
(b) total distance traveled
\(\bigtriangleup 1 + \bigtriangleup 2\)