Question

29) A cheetah can accelerate from rest to 25

m/s in 6 s. Assuming that the cheetah moves
with constant acceleration, what distance does
it cover in the first 3 s

Answer 1

`\huge\boxed{\sf S = 18.75 \ meters}`

Step-by-step explanation:

Given Data:

Initial Velocity = Vi = 0 m/s (rest)

Final Velocity for 6 seconds = Vf = 25 m/s

Time (1) = T1 = 6 seconds

Time (2) = T2 = 3 seconds

Required:

Distance for 3 seconds = S = ?

Solution:

For 6 seconds, the acceleration will be:

`\displaystyle a = (Vf-Vi)/(t) \\\\a = (25 - 0)/(6) \\\\a = 25 / 6\\\\\boxed{a = 4.167 \ m/s^2}`

Since, acceleration is constant, it will be the same at 3 seconds as well.

Using second equation of motion to find Distance (S) with time being 3 seconds:

`\displaystyle S= Vit+(1)/(2) at^2\\\\S = (0)(3)+ (1)/(2) (4.167)(3)^2\\\\S = (1)/(2) (4.167)(9)\\\\S = (37.5)/(2) \\\\\boxed{S = 18.75 \ meters}\\\\\rule[225]{225}{2}`

Hope this helped!

~AH1807