Motion Graphing Acceleration vs Time Graphs acceleration mss
- Slides: 6
Motion Graphing Acceleration vs. Time Graphs • acceleration (m/s/s) • • time (s) If graph has a SLOPE of ZERO; object has a CONSTANT ACCELERATION and VELOCITY is INCREASING. MOTION at a constant acceleration. slope = 0 Vav = increasing a = constant
Motion Graphing Acceleration vs. Time Graphs acceleration (m/s/s) • • time (s) If intervals between acceleration & time intersection points are same; graph represents object moving with INCREASING ACCELERATION; VELOCITY INCREASING greater slope = greater acceleration and velocity is increasing SIGN of slope on an acceleration/time graph indicates CHANGE IN ACCELERATION of object’s travel, or change in DIRECTION of acceleration when below axis POSITIVE (+) slope indicates increasing acceleration, or change in direction of acceleration when below x-axis
Motion Graphing Acceleration vs. Time Graphs acceleration (m/s/s) • • time (s) If intervals between acceleration & time intersection points are same; graph represents object moving with DECREASING ACCELERATION; VELOCITY DECREASING greater slope = greater acceleration and velocity is decreasing SIGN of slope on an acceleration/time graph indicates CHANGE IN ACCELERATION of object’s travel, or change in DIRECTION of acceleration when below x-axis NEGATIVE (-) slope indicates decreasing acceleration, or change in direction of acceleration when below xaxis
Motion Graphing Velocity vs. Time Graphs • velocity (m/s) AREA under curve of a Velocity/Time graph is DISTANCE traveled by object. • Object with initial Constant Velocity Vav =Dx Dx = Vav. Dt Dt time (s)
Motion Graphing Velocity vs. Time Graphs • velocity (m/s) • time (s) AREA under curve of a Velocity/Time graph is DISTANCE traveled by object. Object from rest, velocity increasing at constant rate Dx = 1 a. Dt 2 2
Motion Graphing Velocity vs. Time Graphs • velocity (m/s) • time (s) AREA under curve of a Velocity/Time graph is DISTANCE traveled by object. Object with initial speed, velocity increasing at constant rate Dx = Vi. Dt + 1 a. Dt 2 2
