Acceleration vs Time Primer
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/s 2•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.