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Terms in this set (41)

(1)________________________________________
-the displacement, Δx, can be considered the area under the time interval in a velocity time graph
--since from t = 2 s to t = 3 s there is a slope, we can create "vertical asymptote" at t = 3 s so that we may find the area of a triable on the left side and the area of a rectangle from t= 3 s to t = 4s. The resulting displacement will sum the two areas:
Δx = A1 + A2
-lets make A1 the triangle side and A2 the rectangular side
--we also know that the area of a triangle can be described as A = (1/2)(base)(height)
---for a velocity-time graph, the base is the time along the x-axis and the height is the velocity along the y-axis
---so we have:
A1 = (1/2)(3.0 s - 2.0 s)(0 - 5 m/s)
A1 = (1/2)(1.0 s)(-5 m/s)
A1 = -2.5 m
--the area of a rectangle/square can be described as A = lw
---so we have:
A2 = (4.0 s - 3.0 s)(-5 m/s)
A2 = 1 s(-5 m/s)
A2 = -5 m
-FINALLY:
Δx = A1 + A2
Δx = -2.5 m - 5.0 m
Δx = -7.5 m

(2)________________________________________
Δx = 0.50bh = 0.50vt
= 0.50(1.0 s)(-5.0 m/s)
= -2.5 m

(3)________________________________________
-The area of our velocity vs. time graph is displacement Δx. Returning to the initial position comes at the time where the positive and negative areas cancel out so Δx = 0.
-from the graph, we can see that the velocity becomes constant at t = 3.0 s, so what we are looking for is how much time after t = 3.0s is needed for Δx to = 0?
-let's first find the displacement Δx up to t = 3.0 s, where A1 is from t = 0 s to t = 2.0 s (the first triangle on the left) and A2 is from t = 2.0 s to t = 3.0 s
Δx = A1 + A2
A1 = 0.50(2.0 s)(5.0 m/s) = 5 m/s
A2 = 0.50(1.0 s)(-5.0 m/s) = -2.5 m/s
Δx = 5 m/s + (-2.5 m/s) = 2.5 m/s
-now we want to know how long after t = 3.0 s Δx will = -2.5 m (since Δx = +2.5 m, and we need it to = 0 m)
Δx = A = lw = tv
-2.5 m = t(-5.0 m/s)
t = 0.5 s
-since the question asks at what time we need to add this time after 3.0 s to 3.0s
therefore Δx = 0 at 3.5 s