A sprinter is running a 100m sprint race. The race begins, and stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and pass

Question

A sprinter is running a 100m sprint race. The race begins, and stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, What was the sprinter’s average velocity between the two time marks? Show all your work.

in progress 0
Kim Cúc 3 years 2021-08-07T01:04:16+00:00 1 Answers 41 views 0

Answers ( )

    0
    2021-08-07T01:05:35+00:00

    Answer:

    velocity = 7.7558 m/s

    Explanation:

    s = u × t

    s – distance u – velocity t – time

    59 – 12 = (7.98 – 1.92) *u

    47 = ( 6.06) u

    u = 7.7558 m/s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )