From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.

Question

From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 min, he walks briskly in a straight line at a constant speed of 2.5 m/s.
(a) What is his average velocity vavg in the interval 2.000 min to 9.10 min?
__________m/s

(b) What is his average acceleration aavg in the interval 2.00 min to 9.10 min?
__________ m/s2

(c) What is vavg in time 3.00 min to 9.00 min?
__________ m/s

(d) What is aavg in time 3.00 min to 9.00 min?
__________ m/s2

(e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs. (Do this on paper. Your instructor may ask you to turn in your work.)

Explanation / Answer

(a) Average velocity = S/t  = v.t1/t = [2.5 m/s *(9.1-5.0)*60]/(9.1-2.0)*60= 4.1*2.5/7.1 = 1.444 m/s

(b) Average acceleration =(v-u)/t = 2.5 m/s /7.1min*60 s/min = 2.5/426 = 0.00587 m/s2

(c) V(average) = S/t = 2.5x4/6  = 1.67 m/s

(d) a(average) = Change in v /time = 2.5/6x60 = 0.00694 m/s2

Graph: Draw the v-t graph and S-t graph. Try the rest yourself.

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