A cat is sleeping on the floor in the middle of a 4m wide room when a barking dog enters with a speed of 1.8m/s. As the dog enters, the cat immediately accelerates at 0.85m/s^2 towards an open window on the opposite side of the room. The dog; a bit amazed by the cat began to slow down at 0.2m/s^2 as soon as it enters the room. Do calculations to determine whether the dog will catch the cat before it jumps out of the room via the open window?

Question

A cat is sleeping on the floor in the middle of a 4m wide room when a barking dog enters with a speed of 1.8m/s. As the dog enters, the cat immediately accelerates at 0.85m/s^2 towards an open window on the opposite side of the room. The dog; a bit amazed by the cat began to slow down at 0.2m/s^2 as soon as it enters the room. Do calculations to determine whether the dog will catch the cat before it jumps out of the room via the open window?

Answer 1

The cat jumps out the window before the dog catches it.

Answer 2

To determine whether the dog will catch the cat before it jumps out of the room, we need to calculate the time it takes for both the cat and the dog to reach the window.

First, we calculate the time it takes for the cat to reach the window. We can use the formula:

distance = initial velocity * time + 0.5 * acceleration * time^2

Since the cat starts from rest, its initial velocity is 0, and the distance it needs to cover is the width of the room, which is 4m. Rearranging the equation to solve for time, we get:

time = sqrt(2 * distance / acceleration)

Using the given values, we have:
distance = 4m
acceleration = 0.85m/s^2

Substituting these values into the equation, we find:

time = sqrt(2 * 4m / 0.85m/s^2)
= sqrt(9.41s^2)
≈ 3.07s

So, it will take the cat approximately 3.07 seconds to reach the window.

Next, let's calculate the time it takes for the dog to reach the cat. We can use the formula:

time = (final velocity - initial velocity) / acceleration

Here, the initial velocity of the dog is 1.8m/s, and the dog slows down with an acceleration of -0.2m/s^2. Since the dog is slowing down, its final velocity will be 0m/s when it catches the cat.

Substituting the given values, we have:

time = (0m/s - 1.8m/s) / (-0.2m/s^2)
= (-1.8m/s) / (-0.2m/s^2)
= 9s

So, it will take the dog approximately 9 seconds to catch the cat.

Since the cat reaches the window in 3.07 seconds, and the dog reaches the cat in 9 seconds, we can conclude that the dog will not catch the cat before it jumps out of the room via the open window.