A cheetah can accelerate from rest to 27 m/s in 1.9 s. Assume the acceleration i
ID: 1278336 • Letter: A
Question
A cheetah can accelerate from rest to 27 m/s in 1.9 s. Assume the acceleration is constant over the time interval.
(a) What is the magnitude of the acceleration of the cheetah?
m/s2 in the direction of motion
(b) What is the distance traveled by the cheetah in these 1.9 s?
m
(c) A runner can accelerate from rest to 6.1 m/s in the same time, 1.9 s. What is the magnitude of the acceleration of the runner?
m/s2 in the direction of motion
(d) By what factor is the magnitude of the cheetah's average acceleration greater than that of the runner?
Explanation / Answer
a) a = (v-u)/t = (27-0)/1.9 = 14.21 m/s^2
b) s = 0.5*a*t^2
= 0.5*14.21*1.9^2
= 25.65 m
c)a = (v-u)/t = (6.1-0)/1.9 = 3.21 m/s^2
d) a_cheeta = 14.21 m/s^2
= 14.21*a_person/3.21
= 4.427*a_person
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.