Question

A cat is sleeping on the floor in the middle of a 3.0-m-wide room when a barking dog enters with a speed of 1.50 m/s. As the dog enters, the cat (as only cats can do) immediately accelerates at 0.85 m/s2 toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at 0.10 m/s2 as soon as it enters the room.

How far is the cat in front of the dog as it leaps through the window?

Answer 1

Initial velocity of the cat = V₁ = 0 m/s

Acceleration of the car = a₁ = 0.85 m/s²

The room is 3 m wide and the cat is in the middle of the room.

Width of the room = W = 3 m

Distance of the cat from the window = D₁ = W/2 = 3/2 m = 1.5 m

Time taken by the cat to reach the window = T

D₁ = V₁T + a₁T²/2

1.5 = (0)T + (0.85)T²/2

T = 1.878 sec

Initial velocity of the dog = V₂ = 1.5 m/s

Acceleration of the dog = a₂ = -0.1 m/s² (Negative as it slows down)

Distance traveled by the dog by the time the cat reaches the window = D₂

D₂ = V₂T + a₂T²/2

D₂ = (1.5)(1.878) + (-0.1)(1.878)²/2

D₂ = 2.64 m

At the final position the cat is at the window and the dog is D₂ distance from the opposite end of the window.

Distance between the cat and the dog when the cat leaps out of the window = D

D = W - D₂

D = 3 - 2.64

D = 0.36 m

Distance between the cat and the dog as the cat leaps through the window = 0.36 m