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has length 2.4 s, and the height is v(2.4) = 76.8 ft/s, so the area is
base × height
2
=
2.4 s × 76.8 ft/s
2
= 92.16 ft.
Thus after a mere 2.4 seconds, the ball bearing had already dropped more
than 92 feet!
That solves the specific problem at hand, but let's be more general.
Remember that the larger goal was a kinematic equation x(t) that predicts
an object's position given any initial position and any initial velocity. First,
let's replace the constant 2.4 with an arbitrary time t. Next, let's remove
the assumption that the object initially has zero velocity, and instead allow
an arbitrary initial velocity v
0
. This means the area under the curve v(t)
is no longer a triangle—it is a triangle on top of a rectangle, as shown in
Figure 11.11.
Figure 11.11
Calculating displacement at
time
t
, given initial velocity
v
0
and constant acceleration
a
The rectangle has base t and height v
0
, and its area represents the
distance that would be traveled if there were no acceleration. The triangle
on top of the rectangle also has base t, and the height is at, the difference
in v(t) compared to the initial velocity as a result of the acceleration at the
rate a over the duration of t seconds. Summing these two parts together
yields the total displacement, which we denote as ∆x:
∆x = (Area of rectangle) + (Area of triangle)
+
1
2
Rectangle
base
Rectangle
height
Triangle
base
Triangle
height
=
= (t)(v
0
) + (1/2)(t)(at)
= v
0
t + (1/2)at
2
.
We have just derived a very useful equation, so let's highlight it so that
people who are skimming will notice it.
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