Figure 2.39
We are asked to determine the positionyat various times. It is reasonable to take the initial positiony 0 to be zero. This problem involves one-
dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since
up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, soais
negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration
due to gravity opposes the initial motion and will slow and eventually reverse it.
Since we are asked for values of position and velocity at three times, we will refer to these asy 1 andv 1 ;y 2 andv 2 ; andy 3 andv 3.
Solution for Positiony 1
1. Identify the knowns. We know thaty 0 = 0;v 0 = 13.0 m/s;a= −g= −9.80 m/s^2 ; andt= 1.00 s.
2. Identify the best equation to use. We will usey=y 0 +v 0 t+^1
2
at^2 because it includes only one unknown,y(ory 1 , here), which is the
value we want to find.
3. Plug in the known values and solve fory 1.
y= 0 +(13.0 m/s)(1.00 s)+^1 (2.78)
2
⎛⎝−9.80 m/s
2 ⎞
⎠(^1 .00 s)
(^2) = 8.10 m
Discussion
The rock is 8.10 m above its starting point att= 1.00s, sincey 1 >y 0. It could bemovingup or down; the only way to tell is to calculatev 1
and find out if it is positive or negative.
Solution for Velocityv 1
1. Identify the knowns. We know thaty 0 = 0;v 0 = 13.0 m/s;a= −g= −9.80 m/s^2 ; andt= 1.00 s. We also know from the solution
above thaty 1 = 8.10 m.
2. Identify the best equation to use. The most straightforward isv=v 0 −gt(fromv=v 0 +at, where
a= gravitational acceleration = −g).
- Plug in the knowns and solve.
v (2.79)
1 =v 0 −gt= 13.0 m/s −
⎛⎝9.80 m/s
2 ⎞
⎠(^1 .00 s)= 3.20 m/s
Discussion
The positive value forv 1 means that the rock is still heading upward att= 1.00 s. However, it has slowed from its original 13.0 m/s, as
expected.
Solution for Remaining Times
The procedures for calculating the position and velocity att= 2.00 sand3.00 sare the same as those above. The results are summarized in
Table 2.1and illustrated inFigure 2.40.
Table 2.1Results
Time,t Position,y Velocity,v Acceleration,a1.00 s 8.10 m 3.20 m/s −9.80 m/s^2
2.00 s 6.40 m −6.60 m/s −9.80 m/s^2
3.00 s −5.10 m −16.4 m/s −9.80 m/s^2
Graphing the data helps us understand it more clearly.
CHAPTER 2 | KINEMATICS 63