Asked by Anita
a man run, accelerating at 0.5 m/s^2, for 8 sec. He then slides to a stop in 1.5 sec.
a.) What total distance did he travel?
b.) what was his average velocity?
a.) What total distance did he travel?
b.) what was his average velocity?
Answers
Answered by
Henry
a. d1 = 0.5a*t^2 = 0.25*8^2 = 16 m.
V1 = at = 0.5 * 8=4 m/s=Max. velocity.
a = (V-V1)/t = (0-4) / 1.5=-2.67m/s^2.
d2 = (V^2-V1^2)/2a.
d2=(0-16) / -5.33=3 m.=Sliding distance.
Dt = d1 + d2 = 16 + 3 = 19 m. = Total
dist. traveled.
V1 = at = 0.5 * 8=4 m/s=Max. velocity.
a = (V-V1)/t = (0-4) / 1.5=-2.67m/s^2.
d2 = (V^2-V1^2)/2a.
d2=(0-16) / -5.33=3 m.=Sliding distance.
Dt = d1 + d2 = 16 + 3 = 19 m. = Total
dist. traveled.
Answered by
Henry
b. Vavg = Dt/T = 19 / (8+1.5) = 2m/s.
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