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Sagot :
Answer:
(a) v = 0 m/s
(b) v = 6 m/s
(c) 6 m
(d) a = 0 m/s²
(e) a = 0.75 m/s²
Explanation:
A velocity-time graph shows the velocity (speed) and direction an object travels over a specific period of time.
- y-axis = velocity (in m/s).
- x-axis = time (in seconds).
A horizontal line means constant velocity.
When v = 0 m/s, the object is at rest.
Acceleration is the slope of the line. (A positive slope is acceleration, and a negative slope is deceleration).
Displacement (distance traveled) is the area under the graph.
Part (a)
The object's initial velocity is when t = 0 s.
Therefore, from inspection of the graph, the initial velocity is:
- v = 0 m/s.
Part (b)
The object's final velocity is when t = 8 s.
Therefore, from inspection of the graph, the final velocity is:
- v = 6 m/s
Part (c)
To calculate the displacement of the object from t = 0 s to t = 2 s, find the area under the graph between those times.
The area is a triangle with base 2 and height 3. Therefore, using the area of a triangle formula:
- [tex]\sf Displacement=\dfrac{1}{2} \times 2 \times 3=6\;m[/tex]
Part (d)
The line between t = 2 s and t = 4 s is horizontal. Therefore, the velocity between these times is constant and so the acceleration of the object is zero:
- a = 0 m/s²
Part (e)
To calculate the acceleration of the object from t = 4 s to t = 8 s, find the slope of the line between these two points:
[tex]\implies \textsf{slope}=\dfrac{\textsf{change in $y$}}{\textsf{change in $x$}}=\sf \dfrac{6-3}{8-4}=\dfrac{3}{4}=0.75[/tex]
Therefore, the acceleration of the object from t = 4 s to t = 8 s is:
- a = 0.75 m/s²
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