In order to illustrate the method for solving dvat problems,
consider this problem: A hotair balloon is ascending at the rate
of 12 m/s and is 80 m above the ground when a package is dropped over
the side. How long does it take the package to reach the ground?
(Example 212 in your text is similar to this problem. We provide more
detail about setting the problem up below.)
We'll follow the advice given previously and select the direction of
+x to be up. Note that the diagram indicates this. It also indicates the position of x_{o} and x,
the direction of v_{o,}and the direction of a. The comments in blue are added for explanatory purposes. They would not
be needed in a student solution.
Given: 

x_{o} = 0 
x_{o} is selected to
be zero at the initial position of the package. 
x = 80 m 
The final position is negative, because it's
lower than x_{o}. 
v_{o} = +12 m/s 
The initial velocity is positive, because
it's in the direction of +x. (The balloon is ascending when
the package is released. Hence, the package is initially ascending.) 
v = ? 
The final velocity isn't known. However, we
may not need to know it. 
a = 9.8 m/s² 
The acceleration due to gravity is negative.
Falling objects have increasing negative velocity. Rising objects
have decreasing positive velocity. In either case, the acceleration
is negative. 
Goal: Find the time, t,
for the package to reach the ground. 
Solution: Since the acceleration is uniform,
we can select an appropriate dvat equation. We know
x_{o}, x, v_{o}, and
a, and we need to know t. The dvat with
these 5 variables and no others is x = x_{o} + v_{o}t
+ ½at². 

This is a quadratic equation and can be solved
using the quadratic formula.
Rearrange into standard form and apply the
quadratic formula.
Substitute numerical values with units and
calculate the roots. 
We're interested in the positive root for this problem. Therefore, our answer is t = 5.5 s. (See Example 212
in your text for an interpretation of the negative root.) Checks:
Units: The units of the radical reduce to m/s, the same
as the velocity term in the numerator. The quotient of m/s in the
numerator and m/s² in the denominator is s, as expected.
Sign: We selected the positive root, since time should be
positive.
Sensibility: The package will move a distance of a little
more than 80 m as it first goes up and then back down. 5 seconds is
a reasonable amount of time to go that far under the action of gravity. 
Notice some things we didn't do in solving the problem. We
didn't divide it into two parts, one for the upward motion and another
for downward. There's no need to do that when you select one
direction for +x and use corresponding signs on the given
quantities. Hence, this method saves time.
Something else we didn't do was use g to represent the
acceleration due to gravity. We know that the textbook uses
g, but we're trying to avoid confusion. If you've given
the text a careful reading, you saw that g is defined to
be positive...always. That means if you use g
in a dvat equation with the direction up defined to be
positive, you have to change a (+) sign to a () sign. For
example, x = x_{o} + v_{o}t + ½at²
becomes x = x_{o} + v_{o}t  ½gt².
This creates possibilities for mistakes. Just stick with the
single equation, x = x_{o} + v_{o}t + ½at²,
and you'll avoid mistakes in signs. 
