A cheetah is hunting. Its prey runs for 2.56 s at a constant velocity of +10.02 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

1/2 a 2.56^2=10.02*2.56

25.7

To find the constant acceleration that the cheetah must maintain in order to run the same distance as its prey, we can use the kinematic equation:

𝑑 = 𝑣0𝑡 + (1/2)𝑎𝑡^2,

where 𝑑 is the distance travelled, 𝑣0 is the initial velocity (which is 0 in this case because the cheetah starts from rest), 𝑡 is the time, and 𝑎 is the constant acceleration.

Let's assume that the distance travelled by both the prey and the cheetah is the same. We know the time (2.56 s) and the velocity of the prey (+10.02 m/s). We need to find the acceleration (𝑎).

For the prey:
𝑑 = 𝑣0𝑡 = (10.02 m/s)(2.56 s) = 25.6512 m.

For the cheetah:
𝑑 = (1/2)𝑎𝑡^2 = (1/2)𝑎(2.56 s)^2 = 25.6512 m.

We can solve this equation for 𝑎 by rearranging it:

𝑎 = (2 × 25.6512 m) / (2.56 s)^2 = 10.0415 m/s^2.

Therefore, the cheetah must maintain a constant acceleration of approximately 10.0415 m/s^2 in order to run the same distance as its prey in the same time.