Question

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?

Answer 1

x₁ = 3.481 m

Step-by-step 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 =
`\sqrt{ (12.2)/(0.35) }`

t = 5.9 s

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

x₁ = ½ 0.2 5.9²

x₁ = 3.481 m