To analyze the motion of a brick falling freely from a building, we'll break it down into three graphs: distance-time, speed-time, and acceleration-time.
1. Distance-Time Graph
In free fall, the distance fallen (d) over time (t) can be described by the equation of motion:
\[ d = \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)). Since the brick falls for 5 seconds, you can calculate the distance fallen at \( t = 5 , \text{s} \):
\[ d = \frac{1}{2} \times 9.81 , \text{m/s}^2 \times (5 , \text{s})^2 = \frac{1}{2} \times 9.81 \times 25 = 122.625 , \text{m} \]
The graph will be a parabola opening upwards; it starts at the origin and increases as time goes on, showing that distance increases with the square of time.
2. Speed-Time Graph
The speed (v) of the brick can be calculated using the equation:
\[ v = g t \]
At \( t = 0 , \text{s} \), \( v = 0 , \text{m/s} \), and at \( t = 5 , \text{s} \):
\[ v = 9.81 , \text{m/s}^2 \times 5 , \text{s} = 49.05 , \text{m/s} \]
The graph is a straight line starting from the origin (0,0) and rising linearly to the point (5, 49.05). The slope of the line is constant, reflecting the constant acceleration due to gravity.
3. Acceleration-Time Graph
In free fall, the acceleration (a) is constant and equal to \( g \):
\[ a = 9.81 , \text{m/s}^2 \]
This means the acceleration-time graph will be a horizontal line at \( y = 9.81 , \text{m/s}^2 \) for the entire time of the fall (0 to 5 seconds).
Representation of the Graphs
Distance-Time Graph
Distance (m)
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Time (s)
Speed-Time Graph
Speed (m/s)
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Time (s)
Acceleration-Time Graph
Acceleration (m/s²)
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| g = 9.81
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Time (s)
These graphs illustrate the physics of free fall: distance increases quadratically with time, speed increases linearly, and acceleration remains constant.