http://www.ck12.org Chapter 2. One-Dimensional Motion
suggests that whether we find the triangular area under the graph from 0 to 3 seconds, or the area of a rectangular
using the average speed times the time interval, the result is the same.
FIGURE 2.18
The big triangle shows the rocket slowing
as it goes up. The big rectangle is going
up at the average speed. These have
the same area because the red and blue
areas are the same.This is not a coincidence since 15 m/s is themidpointof thelinearfunctionvf=at+vibetween 0 and 3 seconds.
For the first half of the time interval [0, 1.5] the velocity of the rocket is slower than 15 m/s, and for the second
half of the time interval [1.5, 3.0] it is faster than 15 m/s. The midpoint of the interval can therefore be used as the
average velocity. Thus,vavg=(vi+ 2 vf).
Check Your Understanding
In a circus act, a performer is shot from a vertical cannon with an initial speed of 40 m/s. Using−10 m/s^2 as an
approximation for the acceleration due to gravity:
- Determine the time it takes the performer to reach the highest altitude above the ground.
- Determine the performer’s highest altitude above the ground.
- What is the performer’s displacement att= 7 .0 s (Note, the ground is at position 0.0 m)?
- How far has the performer traveled att= 7 .0 s?
Answer: Though many problems can be solved this way, there are more difficult problems requiring a bit more
mathematical sophistication. We now derive two useful one-dimensional kinematic equations.
All that is necessary in order to derive the next two kinematic equations are the use of:
Equation A:xf=(vi+ 2 vf)t+xiand
Equation B:vf=at+vi
Substitutevffrom B into A and after rearranging and simplifying we have:
xf=^12 at^2 +vit+xi
Using B, findt=vf−avi
Substitutetfrom B into A and after rearranging and simplifying we have:
vf^2 =vi^2 + 2 a∆x