An athlete is training on a 100 m long linear track. His motion is described by

Question

An athlete is training on a 100 m long linear track. His motion is described by the graph of his position vs. time, below. x (m) 100 80 60 40 20 10 20 30 40 50 60 (a) For each segment of the graph, find the magnitude and direction of the athlete's velocity. magnitude VA direction VA magnitude ve direction vB magnitude vc direction vo magnitude VD direction VD m/s Select m/s m/s Select m/s (b) What are the magnitude and direction of the athlete's average velocity over the entire 60 s interval? magnitude direction-Select- m/s positive x negative x Submit Answer Sal The magnitude is zero.

Explanation / Answer

a)

Va= 60/10= 6 m/s

Directed positive x

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Vb= 0 m/s

Magnitude is zero

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Vc= 20/15= 1.33 m/s

Directed negative x

--------

Vd= 60/15= 4 m/s

Directed x axis

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b)

Av velocity= total displacement/time= 100/60= 1.67 m/s

Directed positive x axis

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Comment in case any doubt.. good luck

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