Question

A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards is the positive velocity direction. A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5). A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5). What is the badger's displacement \Delta xΔxdelta, x from t=2\,\text st=2st, equals, 2, start text, s, end text to 3\,\text s3s3, start text, s, end text? Answer with two significant digits. \text mm

Answer 1

The badgers displacement from its position at point t = 2, to its position at point t = 3 is 25 meters

Which method can be used to calculate the displacement of the badger?

The points on the graph are;

(0, 0), (1, 5), (3, -5), (6, -5)

The point t = 2 is between (1, 5) and (3, -5)

The slope, m, of the line containing the point t = 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 t = 2, we have;

v = -5×2 + 10 = 0

The velocity at t = 2 is 0 m/s

The value of the acceleration is the same as the slope, m = a = -5

Given that the initial velocity at t = 2 is 0, we have;

∆x = u•t + 0.5•a•t²

At t = 3, the time of travel from the time t = 2 is 1 seconds, which gives;

∆x = 0×5 + 0.5×(-5)×1² = -2.5

The distance traveled between points, t = 2 and t = 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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