begins to chase the dog with an acceleration of .5 m/s^2. H ow many seconds does it take the chef to catch the dog?
2006-12-10 20:57:55 · 4 answers · asked by Bob Nguyen 1 in Science & Mathematics ➔ Physics
Vd*t + 10 = 1/2 Ac*t^2
Vd = 10 m/s, Ac = 0.5 m/s/s
10 t + 10 = .25 t^2
0 = .25 t^2 - 10 t - 10
quadratic eq gives you t=20 + 2*sqrt(110)
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oops, Vd = 5 m/s
t = 10 + 2*sqrt(35) = about 21.8 s
2006-12-10 21:05:04 · answer #1 · answered by feanor 7 · 0⤊ 1⤋
At the catch, Sd=Sc
10+5*t=1/2*0.5*t^2
t^2-20t-40=0
t=21.8 seconds
2006-12-10 21:26:26 · answer #2 · answered by mekaban 3 · 0⤊ 0⤋
s(chef) = 0.5 * 0.5 * t^2
s(dog) = 10 + 5 * t
s(chef) = s(dog) =>
0.25 * t^2 = 10 + 5 * t =>
t^2 - 20 * t - 40 = 0 =>
(t - 21.83) * (t + 1.83) = 0 =>
t = - 1.83 V t = 21.83
t > 0 => t = 21.83
2006-12-10 21:19:04 · answer #3 · answered by sunnyboy 3 · 0⤊ 0⤋
a 230 grain jacketed hollow point travels about 940fps.
2006-12-10 21:06:36 · answer #4 · answered by glock509 6 · 0⤊ 4⤋