phy asked in Science & MathematicsPhysics · 3 months ago

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

Relevance
  • NCS
    Lv 7
    3 months ago
    Favorite 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!

    Source(s): Thank you for awarding!
Still have questions? Get your answers by asking now.