Abicyclist in the tour de france crests a mountain pass as he moves at 18km/h. at the bottom, 3.6km farther, his speed is 60km/h. what was his average acceleration (in m/s^2) while riding down the mountain?

For this case we can use the following kinematic equation:
 Vf ^ 2 = Vo ^ 2 + 2 * a * d
 Where,
 Vf: final speed.
 Vo: initial speed.
 a: acceleration.
 d: distance.
 We cleared the acceleration:
 Vf ^ 2 = Vo ^ 2 + 2 * a * d
 Vf ^ 2-Vo ^ 2 = 2 * a * d
 a = (Vf ^ 2-Vo ^ 2) / (2 * d)
 Substituting the values:
 a = ((60) ^ 2- (18) ^ 2) / (2 * (3.6))
 a = 455 Km/h^2
 Making change of units:
 a = 455 * (1000 / (3600) ^ 2)
 a = 0.035108025 m / s ^ 2
 
 his average acceleration (in m/s^2) while riding down the mountain is:
 a = 0.035108025 m / s ^ 2

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way