a. a = (Vt - Vo) / t = (24 m/s - 0 m/s) / 2 s = 24 / 2 = 12 m/s^2.
b. d = 0.5at^2 = 0.5 * 12 m/s^2 * (2s)^2 = 0.5 * 12 * 4 = 24 m.
c. a = (6 m/s - 0 m/s) / 2 s 3 m/s^2
12 m/s^2 * ( 1 / 3 )s^2/m = 4.
a cheetah can accelerate from rest to 24 m/s in 2.0 s . Assuming the acceleration is constant over the time interval, a)what is the magnitude of the acceleration of the cheetah? b) what is the distance traveled by the cheetah in these 2.0 s.? c) a runner can accelerate from rest to 6.0 m/s in the same time , 2.0 s What id the magnitude of the acceleration of the runner? By what factor is the cheetah's average acceleration magnitude greater than that of the runner?
2 answers
thank you so much