A jogger maintains a speed of 2.8 m/s for 216 m until he encounters a stoplight,

Question

A jogger maintains a speed of 2.8 m/s for 216 m until he encounters a stoplight, and he abruptly stops and waits 33 s for the light to change. He then resumes his exercise and maintains a speed of 3.8 m/s for the remaining 50 m to his home.

(a) What was his average velocity for this entire time interval?
m/s
(b) What were his maximum and minimum velocities, and how do they compare with this average?

vmax= m/s
vmin= m/s

(c)
(1)The average was closer to the maximum.
(2)The average was closer to the minimum.
(3)The average was halfway between the maximum and minimum.

Explanation / Answer

216/2.8 = 77.123 seconds 50/3.8 = 13.158 seconds So: For 77.123 seconds he maintained a velocity of 2.8 m/s For 13.158 seconds he maintained a velocity of 3.8 m/s For 33 seconds he maintained a velocity of 0 m/s Entire trip lasted 123.281 seconds. 2.8*77.123/123.281 = 1.7516 3.8*13.158/123.281 = .4056 0*33/123.281 = 0 Add them together and you get the average which is 2.1572 m/s The max is 3.8 The min is 0 so: (1)The average was closer to the maximum.

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