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l37645's Question
Physics M
Posted 7 months ago
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The velocity-time graph above represents the motion of a body travelling in a jumpy fashion with periods of constant velocity followed by periods with a different constant velocity.
What is the distance travelled in the interval of time from t=0 st=0 \mathrm{~s} to t=18 st=18 \mathrm{~s} ?
\square m\mathrm{m}

What is the distance travelled in the interval of time from t=4 st=4 \mathrm{~s} to t=8 st=8 \mathrm{~s} ?
\square m

What is the distance travelled in the first 4 seconds? \square m
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Answer from Sia
Posted 7 months ago
Solution
a
To find the distance travelled from t=0st=0 \, \mathrm{s} to t=18st=18 \, \mathrm{s}, calculate the area under the velocity-time graph for this interval
b
For the interval from t=4st=4 \, \mathrm{s} to t=8st=8 \, \mathrm{s}, calculate the area under the velocity-time graph between these times
c
To find the distance travelled in the first 4 seconds, calculate the area under the velocity-time graph from t=0st=0 \, \mathrm{s} to t=4st=4 \, \mathrm{s}
Answer
[Insert final answer here for each question after calculating the areas]
Key Concept
The area under a velocity-time graph represents the displacement of the object over that time interval. If the velocity is constant, the area is a rectangle; if the velocity changes, the area may be a trapezoid or other shape.
Explanation
To find the distance travelled, we look at the shape formed under the velocity-time graph for the given time intervals and calculate the area of those shapes.


Repeat this format for each question the student has posed.

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