http://www.ck12.org Chapter 2. One-Dimensional Motion
FIGURE 2.14
squared. However, the area in the acceleration-time plane is different because thexandyaxes do not have units of
length. They−axis has units of acceleration, m/s^2 , and thex−axis has units of time, s; the area, therefore, has units
of
(m
s^2)
×
(s
1)
=ms. But m/s are the units for either speed or velocity. If we treat acceleration as a vector, allowing
for a change in direction, then the area in the acceleration-time plane is the change in velocity. Using the area to
compute speed from an acceleration-time graph is only correct for motion in one direction. However, computing the
change in velocity from an acceleration-time graph is always true, and so unless otherwise stated, we will only use
this method for velocity calculations.
Use the area under an acceleration-time graph to create a velocity-time table. Note the velocity is the instantaneous
velocity.
Consider the rectangular area formed for the time interval [0, 1] (horizontal “length”) and the corresponding accel-
eration interval, [0, 2] (vertical “length”). The area for the interval is( 1 .0 s)×( 2 .0 m/s^2 ) = 2 .0 m/s. The result is
interpreted, as follows: During the time interval [0, 1] the car has gained a velocity of 2.0 m/s. Thus, at the instant
t= 1 .0 s, the car has an instantaneous velocity of 2.0 m/s. The area of the rectangle for the time interval [0, 2] and
the corresponding acceleration interval [0, 2]= ( 2 .0 s)×( 2 .0 m/s^2 ) = 4 .0 m/s. Therefore, att= 2 .0 s, the car has
an instantaneous velocity of 4.0 m/s. Repeated applications of this process for time intervals [0, 3] and [0, 4], leads
to theTable2.2.
TABLE2.2: Velocity-Time Table
t(s) v(m/s)
0.0 0.0
1.0 2.0
2.0 4.0
3.0 6.0
4.0 8.0FIGURE 2.15
Notice that for any interval of 1.0 s, the car gains another 2.0 m/s of velocity, which is exactly what is meant by
an acceleration of 2.0 m/s^2 → 2 .0 (m/s)/s→ 2 .0 m/smore, with each passing second, s. This suggests an equation
to compute instantaneous velocity using uniform acceleration: Vf=at. Ifa= 2 .0 m/s^2 , then,V= 2 t, thus, at
t= 3 .0 s,V= 2. 0 ( 3. 0 ) = 6 .0 m/s, just as in the table. (Note:tin the equation is any real number. Whent= 1 .2 s,
the instantaneous velocity is 2.4 m/s.)
The equationvf=atshould not come as a surprise if we consider the definition of acceleration.a=∆∆tv=((vtff−−vtii)).
Rearranging we have:(vf−vi) =a(tf−ti). Assuming thatti=0, we have,vf=at+vi.