Question
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
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.
Henry
b. Vavg = Dt/T = 19 / (8+1.5) = 2m/s.