Question

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)

Answer 1

Answer:

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