The skier begins to descend from the mountain, accelerating with an acceleration of 0.5

The skier begins to descend from the mountain, accelerating with an acceleration of 0.5 m / s squared for 10 s, and then rolls along the plain for a while until a complete stop with an acceleration of 0.2 m / s squared. How many meters does the skier cover during the entire movement?

V0 = V = 0 m / s.

a1 = 0.5 m / s2.

t1 = 10 s.

a2 = 0.2 m / s2.

S -?

The entire path S traversed by the skier while moving will be the sum: S = S1 + S2, where S1 is the downhill path, S2 is the skier’s path along the plain.

With uniformly accelerated downhill movement, we express the path by the formula: S1 = V0 * t1 + a1 * t1 ^ 2/2. Since by the condition of the problem V0 = 0 m / s, then S1 = a1 * t1 ^ 2/2.

S1 = 0.5 m / s2 * (10 s) ^ 2/2 = 25 m.

The speed of the skier at the end of the descent and the beginning of the horizontal movement V1 is expressed by the formula: V1 = √ (2 * S1 * a1).

V1 = √ (2 * 25 m * 0.5 m / s2) = 5 m / s.

S2 = (V1 ^ 2 – V2) / 2 * a2.

Since at the end of the movement along the plain the skier stopped V = 0 m / s, then S2 = V1 ^ 2/2 * a2.

S2 = (5 m / s) ^ 2/2 * 0.2 m / s2 = 62.5 m.

S = 62.5 m + 25 m = 87.5 m.

Answer: the skier’s path was S = 87.5 m.