Question

A cheetah is walking at 5.0m/s when it sees a zebra 35m away. What acceleration would be required for the cheetah to reach 25.0 m/s in that distance?

Answer 1

Answer:

Acceleration, $a=8.57\ m/s^2$

Explanation:

Given that,

Initial speed of the Cheetah, u = 5 m/s

Distance covered, s = 35 m

Final speed of the cheetah, v = 25 m/s

We need to find the acceleration required for the cheetah to reach its final speed. The formula is as follows :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(25)^2-(5)^2}{2\times 35}$

$a=8.57\ m/s^2$

So, the acceleration of the cheetah is $8.57\ m/s^2$. Hence, this is the required solution.