Graphical Analysis

Question 1

Graphical analysis:

Finding Velocity from Position-Time Graph

Answer 1

1. Identify two points on the position-time graph.
   Calculate the slope:
   𝑣=Δ𝑦/Δ𝑥

Question 2

Example: If the position changes from 2 m to 6 m over 2 seconds,

Answer 2

𝑣=6−2/2−0
=2m/s

Question 3

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.

Answer 3

position-time graph

Question 4

In a position-time graph the slope represents the object’s _____. A straight line indicates constant ___, while a curved line indicates ___$.

Answer 4

Slope= Velocity
Straight line= Constant velocity
Curved lines = Acceleration

Question 5

A ____ graph shows an object’s velocity relative to time. The slope represents acceleration. The area under the curve represents the object’s displacement

Answer 5

velocity-time

Question 6

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.

Answer 6

Slope= Acceleration
Area under curve = Displacement

Question 7

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.

Answer 7

Slope= Acceleration
Area under curve = Displacement

Question 8

Finding Displacement from Velocity-Time Graph

Answer 8

1. Determine the area under the velocity-time graph.
   For a rectangular area:
   Displacement=Velocity×Time

Question 9

Example: If velocity is 3 m/s over 4 seconds

Answer 9

Displacement=3×4=12m

Question 10

An _____ graph shows an object’s acceleration relative to time. The area under the curve represents the change in velocity

Answer 10

acceleration-time graph

Question 11

In acceleration-time graph the area under the curve represents the change in ___.

Answer 11

Area under curve = change in velocity

Question 12

Finding Change in Velocity from Acceleration-Time Graph

Answer 12

1. Determine the area under the acceleration-time graph.
2. For a rectangular area: Δ𝑣=𝑎×𝑡
   .

Question 13

Example: If acceleration is 2 m/s² over 5 seconds,
.

Answer 13

Δ𝑣=2x5
=10m/s

Question 14

Curved lines on a position-time graph indicate changing velocity (acceleration). Concave up indicates ___ acceleration, concave down indicates ___ acceleration.

Answer 14

Up- positive
Down - negative

Question 15

Curved lines on a position-time graph indicate changing velocity (acceleration). Concave up indicates ___ acceleration, concave down indicates ___ acceleration.

Answer 15

Up- positive
Down - negative

Question 16

On a position-time graph, uniform motion is represented by a ___ line. On a velocity-time graph, it is represented by a ___ line.

Answer 16

UNIFORM MOTION
Position-time - Straight
Velocity- time - Horizontal

Question 17

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.

Answer 17

UNIFORMLY ACCELERATED MOTION
Position-time - Parabolic
Velocity-time Straight

Question 18

Example: Uniformly Accelerated Motion (Position-Time)

If initial velocity 𝑢=0, acceleration 𝑎=3m/s², and time 𝑡=4s

Answer 18

1. 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

Question 19

On a position-time graph, a straight, ___ line represents constant velocity. On a velocity-time graph, a __ line indicates constant velocity.

Answer 19

CONSTANT VELOCITY
PT - Diagonal
VT- Horizontal

Question 20

Example: Constant Velocity (Velocity-Time)

If velocity is 5 m/s for 6 seconds

Answer 20

1. The graph is a horizontal line.
2. Displacement is the area under the graph:
   Displacement=Velocity×Time

Hence,
Displacement=5×6=30m.

Question 21

Example: Constant Velocity (Velocity-Time)

If velocity is 5 m/s for 6 seconds

Answer 21

The graph is a horizontal line.
Displacement is the area under the graph:
Displacement=Velocity×Time

Question 22

On a position-time graph, a ___ line represents constant acceleration. On a velocity-time graph, a straight, ___ line indicates constant acceleration.

Answer 22

CONSTANT ACCELERATION
PT- Curved
VT- sloped

Question 23

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:

Answer 23

1. Break the graph into shapes (rectangles, triangles).
2. 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

Question 24

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:

Answer 24

1. Break the graph into shapes (rectangles, triangles).
2. 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