Asked by annonomus
What distance is travelled by a basketball player who accelerates from rest to 5.654x10^0 m/s in 1.567x10^0 seconds, to three significant figures?
Can someone please explain how to do the calculations
Can someone please explain how to do the calculations
Answers
Answered by
Henry
a = (Vt - Vo) / t,
a = (5.654 - 0) m/s / 1.567 s,
a = 5.654 / 1.567 = 3.61 m/s^2.
d = 0.5at^2 = 0.5*3.61*(1.567)^2 =
4.43 m.
a = (5.654 - 0) m/s / 1.567 s,
a = 5.654 / 1.567 = 3.61 m/s^2.
d = 0.5at^2 = 0.5*3.61*(1.567)^2 =
4.43 m.
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