Answer:
Refer below.
Explanation:
We are given a velocity vs. time graph representing the motion of some object. We are asked to describe the motion of the object between 0 s < t < 8 s.
[tex]\hrulefill[/tex]
We can determine two things by looking at a velocity vs. time graph:
Slope of a Velocity-Time Graph
The slope of a velocity-time graph indicates the acceleration of the object. Here's how it's defined:
- Positive Slope: If the slope is positive (i.e., the line inclines upwards as it moves from left to right), the object is accelerating.
- Negative Slope: If the slope is negative (i.e., the line declines as it moves from left to right), the object is decelerating.
- Zero Slope: If the slope is zero (i.e., a horizontal line), the object maintains a constant velocity.
Area Under a Velocity-Time Graph
The area enclosed between the velocity-time curve and the time axis represents the displacement of the object over that time interval. Here’s how it works:
- Above the Time Axis: If the curve is above the time axis, the area corresponds to positive displacement (movement in the positive direction).
- Below the Time Axis: If the curve is below the time axis, the area corresponds to negative displacement (movement in the negative direction).
- Mixed Above and Below: If the curve crosses the time axis, parts of the area can be positive and parts can be negative, which can indicate changes in the direction of movement.
[tex]\hrulefill[/tex]
Now let's analyze the graph:
Looking at the given graph, there are four distinct changes in the velocity of the object. These occur when:
- 0 s < t < 1 s
- 1 s < t < 5 s
- 5 s < t < 6 s
- 6 s < t < 8 s
Let's look at each time frame.
When 0 s < t < 1 s:
[tex]\text{Slope} = \text{Acceleration} =\dfrac{\Delta y}{\Delta x} = \dfrac{\Delta \vec v}{\Delta t}= \dfrac{8-0 \text{ m/s}}{1-0 \text{ s}}= \boxed{8 \text{ m/s}^2}[/tex]
[tex]\text{Area} = \Delta \vec x=\dfrac{1}{2}bh=\dfrac{1}{2}(1 \text{ s})(8 \text{ m/s})=\boxed{4 \text{ m}}[/tex]
Thus, the object was accelerating at 8 m/s² and displaced 4 meters.
When 1 s < t < 5 s:
[tex]\text{Slope} = \text{Acceleration} =\dfrac{\Delta y}{\Delta x} = \dfrac{\Delta \vec v}{\Delta t}= \dfrac{8-8 \text{ m/s}}{5-1 \text{ s}}= \boxed{0 \text{ m/s}^2}[/tex]
[tex]\text{Area} = \Delta \vec x=bh=(4 \text{ s})(8 \text{ m/s})=\boxed{32 \text{ m}}[/tex]
Thus, the object was moving at constant velocity and displaced 32 meters.
When 5 s < t < 6 s:
[tex]\text{Slope} = \text{Acceleration} =\dfrac{\Delta y}{\Delta x} = \dfrac{\Delta \vec v}{\Delta t}= \dfrac{4-8 \text{ m/s}}{6-4 \text{ s}}= \boxed{-4 \text{ m/s}^2}[/tex]
[tex]\text{Area} = \Delta \vec x=\dfrac{1}{2}bh+bh=\dfrac{1}{2}(1 \text{ s})(4 \text{ m/s})+(1 \text{ s})(4 \text{ m/s})=\boxed{6 \text{ m}}[/tex]
Thus, the object was decelerating at 4 m/s² and displaced 6 meters.
When 6 s < t < 8 s:
[tex]\text{Slope} = \text{Acceleration} =\dfrac{\Delta y}{\Delta x} = \dfrac{\Delta \vec v}{\Delta t}= \dfrac{4-4 \text{ m/s}}{8-6 \text{ s}}= \boxed{0 \text{ m/s}^2}[/tex]
[tex]\text{Area} = \Delta \vec x=bh=(2 \text{ s})(4 \text{ m/s})=\boxed{8 \text{ m}}[/tex]
Thus, the object was moving at constant velocity and displaced 8 meters.
The total displacment is the sum of the area under the entire graph, thus:
[tex]\Delta \vec x_{\text{tot.}}=4 +32+6+8 \text{ m}=\boxed{ 50 \text{ m}}[/tex]
