The badgers displacement from its position at point <em>t </em>= 2, to its position at point <em>t </em>= 3 is 25 meters
<h3>Which method can be used to calculate the displacement of the badger?</h3>
The points on the graph are;
(0, 0), (1, 5), (3, -5), (6, -5)
The point <em>t </em>= 2 is between (1, 5) and (3, -5)
The slope, <em>m</em>, of the line containing the point <em>t </em>= 2 is therefore;
- m = (-5 - 5)/(3 - 1) = -5
The equation of the line is therefore;
v - 5 = -5•(t - 1)
v = -5•t + 5 + 5 = -5•t + 10
v = -5•t + 10
At <em>t </em>= 2, we have;
v = -5×2 + 10 = 0
The velocity at <em>t </em>= 2 is 0 m/s
The value of the acceleration is the same as the slope, <em>m </em>= a = -5
Given that the initial velocity at <em>t </em>= 2 is 0, we have;
∆x = u•t + 0.5•a•t²
At <em>t </em>= 3, the time of travel from the time <em>t </em>= 2 is 1 seconds, which gives;
∆x = 0×5 + 0.5×(-5)×1² = -2.5
The distance traveled between points,<em> </em><em>t </em>= 2 and <em>t </em>= 3 is ∆x = -2.5, which is 2.5 meters in the direction opposite to the initial direction.
Learn more about the equations of motion here:
brainly.com/question/21010554
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