Acheetah is walking at 1.0 m/s when it sees a zebra 25 m away. what acceleration would be required to reach 20.0 m/s in that distance?

Let
 a = acceleration.
 v = Speed
 r = position
 t = time.
 To find the answer, you must write the equations of speed, and position known and clear the variables that are needed to find the acceleration.
 The answer is a = 7.98m / s ^ 2.
 I attach the solution.