Cheetahs, the fastest of the great cats, can reach 45 mph in 2.0 sec starting from rest. Assuming that they have constant acceleration throughout that time, find (a) their acceleration (in f t/s2 and m/s2) (b) the distance they travel during that time (in m and ft)

acceleration is 10.05 m/s² or 32.97 ft/s²

distance is 40.22 m or 131.95 ft

Explanation:

given data

velocity = 45 mph = 20.1168 m/s

time = 2 sec

to find out

acceleration and distance

solution

first we will apply here formula for acceleration that is

v = u + at ... 1

here v is velocity and u is initial speed and t is time and a is acceleration

so put here all value

20.11 = 0 + a (2)

a = 10.05 m/s

so acceleration is 10.05 m/s² or 32.97 ft/s²

and

distance is calculated as

distance = v * t ... 2

put here value

distance = 20.11 * 2

distance = 40.22

so distance is 40.22 m or 131.95 ft