Cheetahs can accelerate to a speed of 21.3 m/s in 2.70 s and can continue to accelerate to reach a top speed of 29.9 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +? direction point in the direction the cheetah runs.
Express the cheetah's top speed ?topvtop in miles per hour (mi/h).
Starting from a crouched position, how much time ?acceltaccel does it take a cheetah to reach its top speed and what distance ? does it travel in that time?
If a cheetah sees a rabbit 138.0 m away, how much time ? total will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?
v1 = at1
a= v1/t1 = 21.3/2.7 = 7.88 m/s^2
(a)
1 m/s = 2.23693629 miles / hour
v top = 29.9 ( 2.23693629 miles / hour) = 66.88 mi/h
(b)
v= u+ at
u = 0
t = v/a = 29.9/7.88 = 3.79 s
(c)
the distance travelled in time t is
s= ut + 1/2 at^2 = 0+1/2 *7.88 ( 3.79)^2 =56.59 m
the remaining distance is
s' = 138-56.59=81.4 m
t' = s'/vtop = 81.4 m/29.9 m/s = 2.72 s
total time
t total = t+ t' = 3.79 s +2.72 s = 6.51 s
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