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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 $t=0 \mathrm{~s}$ to $t=18 \mathrm{~s}$ ?
$\square$ $\mathrm{m}$
What is the distance travelled in the interval of time from $t=4 \mathrm{~s}$ to $t=8 \mathrm{~s}$ ?
$\square$ 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 $t=0 \, \mathrm{s}$ to $t=18 \, \mathrm{s}$, calculate the area under the velocity-time graph for this interval

b

For the interval from $t=4 \, \mathrm{s}$ to $t=8 \, \mathrm{s}$, 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 $t=0 \, \mathrm{s}$ to $t=4 \, \mathrm{s}$

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.

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