91Ó°ÊÓ

The speed of the dog is $v_D=1.5m/s$$v_D=1.5m/s$

The acceleration of a cat is $a_c=0.85m/s^2$

The acceleration of the dog after entering the room is $a_D=-0.10m/s^2$

The width of the room is $S=3m$

Let the initial speed of the cat is $u_C=0m/s$
The distance between the cat and the dog as soon as it enters the room is $s=1.5m$

Write the equation of motion for the cat

$$\begin{gathered} s=u_{c}t+\frac{1}{2}{a}_c{t}^2 \\ 1.5=0+\frac{1}{2}(0.85)t^2 \\ t=\sqrt{\frac{3}{0.85}} \\ t=1.87sec \end{gathered}$$

Thus, the time taken by the cat to reach the window is 1.87 sec.

Write the equation of motion for the dog to determine the distance it covered in time of 1.87 sec

$$\begin{gathered} s=v_{D}t+\frac{1}{2}{a}_D{t}^2 \\ s-\left(1.5\right)(1.87)+\frac{1}{2}(-0.10){(1.87)}^2 \\ s=2.8-0.17=2.63 \\ s=2.63m \end{gathered}$$

Thus, the dog is at a distance of 2.63 m from the entrance door of the room.
The cat is already at the window, this implies that it is 3m away from the door.
Therefore, the distance between the dog and the cat is $3-2.63=0.37m$

Therefore, the dog would not be able to catch the cat and they are 0.37 m apart from each other.