# Physics . this is so hard?

From t = 0 to t = 4.38 min, a man stands still, and from t = 4.38 min to t = 8.76 min, he walks briskly in a straight line at a constant speed of 1.59 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 5.38 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 6.38 min?

### 1 Answer

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- NCSLv 71 month agoFavorite Answer
(a) During the specified interval, he covered

d = (5.38 - 4.38)*60s * 1.59m/s = 95.4 m

and so his average velocity was

Vavg = 95.4m / (5.38-1.00)*60s = 0.363 m/s

and acceleration

a = Δv / Δt = 1.59m/s / (5.38-1.00)*60s = 0.00605 m/s²

(b) Now he's covered

d = (6.38 - 4.38)*60s * 1.59m/s = 191 m

Vavg = 191m / (6.38 - 2.00)*60s = 0.726 m/s

a = 1.59m/s / (6.38 - 2.00)*60s = 0.00605 m/s²

Hope this helps!

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