Graphical Analysis Flashcards
Graphical analysis:
Finding Velocity from Position-Time Graph
- Identify two points on the position-time graph.
Calculate the slope:
𝑣=Δ𝑦/Δ𝑥
Example: If the position changes from 2 m to 6 m over 2 seconds,
𝑣=6−2/2−0
=2m/s
shows an object’s position relative to time. The slope represents the object’s velocity. A straight line indicates constant velocity, while a curved line indicates acceleration.
position-time graph
In a position-time graph the slope represents the object’s _____. A straight line indicates constant ___, while a curved line indicates ___$.
Slope= Velocity
Straight line= Constant velocity
Curved lines = Acceleration
A ____ graph shows an object’s velocity relative to time. The slope represents acceleration. The area under the curve represents the object’s displacement
velocity-time
In velocity-time graph shows an object’s velocity relative to time. The slope represents ____. The area under the curve represents the object’s displacement.
Slope= Acceleration
Area under curve = Displacement
In velocity-time graph shows an object’s velocity relative to time. The slope represents ____. The area under the curve represents the object’s displacement.
Slope= Acceleration
Area under curve = Displacement
Finding Displacement from Velocity-Time Graph
- Determine the area under the velocity-time graph.
For a rectangular area:
Displacement=Velocity×Time
Example: If velocity is 3 m/s over 4 seconds
Displacement=3×4=12m
An _____ graph shows an object’s acceleration relative to time. The area under the curve represents the change in velocity
acceleration-time graph
In acceleration-time graph the area under the curve represents the change in ___.
Area under curve = change in velocity
Finding Change in Velocity from Acceleration-Time Graph
- Determine the area under the acceleration-time graph.
- For a rectangular area: Δ𝑣=𝑎×𝑡
.
Example: If acceleration is 2 m/s² over 5 seconds,
.
Δ𝑣=2x5
=10m/s
Curved lines on a position-time graph indicate changing velocity (acceleration). Concave up indicates ___ acceleration, concave down indicates ___ acceleration.
Up- positive
Down - negative
Curved lines on a position-time graph indicate changing velocity (acceleration). Concave up indicates ___ acceleration, concave down indicates ___ acceleration.
Up- positive
Down - negative
On a position-time graph, uniform motion is represented by a ___ line. On a velocity-time graph, it is represented by a ___ line.
UNIFORM MOTION
Position-time - Straight
Velocity- time - Horizontal
On a position-time graph, uniformly accelerated motion is a ____ curve. On a velocity-time graph, it is a ___ line with a non-zero slope.
UNIFORMLY ACCELERATED MOTION
Position-time - Parabolic
Velocity-time Straight
Example: Uniformly Accelerated Motion (Position-Time)
If initial velocity 𝑢=0, acceleration 𝑎=3m/s², and time 𝑡=4s
- The graph is a parabola (curved line).
2.Use the equation s=𝑢𝑡+1/2x 𝑎 x 𝑡² to calculate position.
Hence;
s=0+1/2 ×3×4²
=24m
On a position-time graph, a straight, ___ line represents constant velocity. On a velocity-time graph, a __ line indicates constant velocity.
CONSTANT VELOCITY
PT - Diagonal
VT- Horizontal
Example: Constant Velocity (Velocity-Time)
If velocity is 5 m/s for 6 seconds
- The graph is a horizontal line.
- Displacement is the area under the graph:
Displacement=Velocity×Time
Hence,
Displacement=5×6=30m.
Example: Constant Velocity (Velocity-Time)
If velocity is 5 m/s for 6 seconds
The graph is a horizontal line.
Displacement is the area under the graph:
Displacement=Velocity×Time
On a position-time graph, a ___ line represents constant acceleration. On a velocity-time graph, a straight, ___ line indicates constant acceleration.
CONSTANT ACCELERATION
PT- Curved
VT- sloped
Example: Variable Acceleration (Acceleration-Time)
For a rectangle with height 2 m/s² and width 3 s, and a triangle with height 2 m/s² and base 3 s:
- Break the graph into shapes (rectangles, triangles).
- Calculate the area of each shape and sum them up.
Hence,
Rectangle area:
2×3
=6m/s
Triangle area:
1/2 ×2×3
=3m/s
Total change in velocity:
6+3=9m/s
Example: Variable Acceleration (Acceleration-Time)
For a rectangle with height 2 m/s² and width 3 s, and a triangle with height 2 m/s² and base 3 s:
- Break the graph into shapes (rectangles, triangles).
- Calculate the area of each shape and sum them up.
Hence,
Rectangle area:
2×3
=6m/s
Triangle area:
1/2 ×2×3
=3m/s
Total change in velocity:
6+3=9m/s