Shown below are six graphs of velocity versus time during a 4 seconds period for

Question

Shown below are six graphs of velocity versus time during a 4 seconds period for six identical objects. The objects move along a straight line on a horizontal surface. This figure will be used for problems 1-3. For the six cases shown above rank the situations based on the displacement that occurs during the 4 seconds from greatest to least. Rank negative numbers lower than zero and positive values. Example: X=+2 m, Y=0 m, Z=-2 m would rank X=1, Y=2, Z=3 For the six cases shown above rank the situations based on the average velocity during the 4 seconds from greatest to least. Rank negative numbers lower than zero and positive values. Example: X=+2 m/s, Y=0 m/s, Z=-2 m/s would rank X=1, Y=2, Z=3 For the six cases shown above rank the situations based on the acceleration that occurs during the 4 seconds from greatest to least. Rank negative numbers lower than zero and positive values. Example: X=+2 m/s^2, Y=0 m/s^2, Z=-2 m/s^2 would rank X=1, Y=2, Z=3

Explanation / Answer

1) Displacement 'd' is equal to area bounded curve and the velocity and time axes.

a) d1 = (1/2)4.2 = +4m, d2 = (1/2)4(-2) = -4m, d3 = (1/2)(-1)(2)+(1/2)(1)(2) = 0m, d4 =(1/2)(2)(2)+(1/2)(2)(-2) = 0m , d5 = (1/2)4.2 =+4m, d6 = (1/2)(2)(1)+(1/2)(2)(-1) = 0m

Hence, A = E = 1, C=D=F = 2, B = 3

b) velocity = displacement/time

v1 = 4/4 = 1m/s, v2 = -1m/s, v3 = 0, v4 = 0, v5 = 1m/s, v6 = 0

Hence, A = E = 1, C=D=F = 2, B = 3

c) acceleration =Slope of the graph

a1 = 2/4 = 1/2 m/s^2, a2 = -1/2 m/s^2, a3 = -1-1/4 = 1/2m/s^2, a4 = 2-(-2)/4 = 1m/s^2, a5 = 1/2 m/s^2, a6 = 1-(-1)/4 = 1/2 m/s^2

Hence, D = 1, A = C =E = F = 2, B = 3

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