| An Acceleration vs Time graph lends itself to a more restricted analysis than does a Displacement vs Time or a Velocity vs Time graph. What information |
| can be determined by analyzing the graph below which is the same graph drawn by the animator? |
| |
|
|
| |
| Qualitatively, one can look at the graph and determine the rate of change of acceleration or deceleration (instantaneous acceleration). This graph shows |
| a constant deceleration for the entire 4.0 s interval. As the projectile is moving upward in a positive direction, it is constantly decreasing in velocity at the |
| rate of -9.80 m/s2. During its downward flight, the projectile is increasing in velocity but in a negative direction. |
| |
| The area between the graph and the time axis equals the object’s change in velocity (Δv). The area shaded blue above the x-axis will be positive |
| because both the acceleration and the time are greater than zero. The area shaded blue below the x-axis will be negative because the acceleration is |
| negative and the time is greater than zero. |
| |
|
|
| |
| To determine the Δv from 0 to 4.00 s, you would determine the area of a rectangle: |
| |
A = Δv = b•h = a•Δt = -9.80 m/s2•4.0 s = -39.2 m/s |
| |
| This is easy to remember starting with the kinematics formula: vf = vi + aΔt |
| vf – vi = aΔt |
| Δv = aΔt |
| |
| This answer also agrees with the result that could have been determined from the data table. |
| |
| Don’t overlook the y-intercept because it tells you the initial conditions (when the observer decided to start collecting data). In either graph above, at |
| t = 0, a = -9.80 m/s2. Keep in mind that the initial value of “t” does not have to equal zero. |
