http://www.ck12.org Chapter 2. One-Dimensional Motion
FIGURE 2.16
We can generate a velocity-time table in the same way as before by determining the area in the acceleration-time
plane or by using the equation:vf=at+vi. If we use the area method we see after one second the area is -10
m/s. The negative sign implies that after the first second of travel, the object has lost 10 m/s of its speed. Since
the initial velocity was +30 m/s, it has +20 m/s of velocity remaining by the end of the first second of travel. Since
the acceleration-time graph is horizontal, each proceeding second of time will again have an area of -10 m/s. Thus,
as each second goes by, the rocket loses another 10 m/s of velocity. The equationvf=− 10 t+30 shows this very
nicely.
See theTable2.3:
TABLE2.3:
T(s) V(m/s)
0 +30
1 +20
2 +10
3 0
4 -10
5 -20
6 -30FIGURE 2.17
There are two very important questions we can ask as we consider this graph.
- Do the units of the slope provide any information concerning the rocket?
Answer:Yes. The units of the slope are m/s divided by s, which equal m/s^2. Thus, the slope has units of acceleration.
The numerical value of the slope (choose any two points on the line) is−10 m/s^2. Now that we know the slope,
m=−10 m/s^2 , let’s write the equation of the line. In a pure math class, the slope of a line is represented by
y=mx+b. In this case, however, our vertical axis isVfor velocity while our horizontal axistfor time. The
constantmis the slope, that we will call “a” for acceleration. The constantbis the vertical intercept, which we will
callvifor velocity (initial), meaning the velocity at timet=0. This gives us
v=at+vi
If we replace our variable names for the specific values in this problem, the resulting equation is:
v=−10 m/s^2 ×t+30 m/s