A dog and a sledge are on the frictionless ice of a frozen lake, 12.2 m apart but connected by a rope of negligible mass. The dog exerts a certain horizontal force (N) on the rope. If
magnitudes of the sledge and the dog accelerations are 0.5 m/s2 and 0.2 m/s2,
respectively. How far from the dog's initial position (m) do they meet?

    x₁ = 3.481 m

Explanation:

In this exercise the acceleration of the dog and the sled are given, for which we can use the kinematic relations.

Let's start by fixing a reference system located on the sled and suppose that it moves to the right, the sled acceleration a₁ = 0.2 m / s, the dog is on the other side of the rope, so if the acceleration is a₂ = -0.5 m / s2, the negative sign indicates that it is moving to the left.

The initial position of the sled is x₁₀ = 0 and the initial position of the dog is x₂₀ = 12.2 m, let's write the expressions for the position

            x₁ = x₁₀ + v₁₀ t + ½ a₁ t²

            x₂ = x₂₀ + v₂₀ t + ½ a₂ t²

in this case the two bodies start from rest so their initial velocities are zero

we substitute the values

           x₁ = 0 + 0 + ½ (0.2) t²

           x₂ = 12.2 + 0 + ½ (-0.5) t²

at the point where the two meet, the position is the same

           x₁ = x₂

          ½ 0.2 t² = 12.2 - ½ 0.5 t²

          (0.1 + 0.25) t² = 12.2

                t =

                t = 5.9 s

we look for the position of the sled (subscript 1)

            x₁ = ½  0.2  5.9²

            x₁ = 3.481 m

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min