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Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.

a.) assuming that they start at the same point ant the same time, how much sooner does the faster arrive at a destination 780 m away?

b.) how far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

  • Physics -

    (a) distance = speed x time
    or time = distance/speed
    The two times can be found by dividing the distance by each speed. Subtract the two times.

    (b) Let t = time it takes the slower student.
    5.50 min = 330 s
    The distance is
    0.90t = 1.90(t-330)
    Solve for t, then substitute into 0.90t

  • Physics -

    B. 570m

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