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/s 2 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/s 2 as soon as it enters the room. does the dog catch the cat before the cat is able to leap through the window?

Answer 1

Good morning.


Lets make the movement function for the dog and cat.

The cat has a start position of1.5 m(the middle of the room), with an initial speed of 0 and acceleration of 0.85 m/s².

The dog has a start position of 0, an initial speed of 1.5 m/s and acceleration of -0.1 m/s².


Cat:

$\mathsf{X = X_0+V_0t + \dfrac{at^2}{2}}\\ \\ \mathsf{X = 1.5 + \dfrac{0.85t^2}{2}}\\ \\ \\ \mathsf{X_c = 1.5 + 0.425t^2}$

Dog:

$\mathsf{X_d= 1.5t - 0.05t^2}$


Let's see if the dog reach the cat. Physically, it means $\mathsf{X_d = X_c}$

$\mathsf{1.5t - 0.05t^2 = 1.5 + 0.425t^2}\\ \\ \mathsf{0.425t^2 + 0.05t^2 - 1.5t + 1.5 = 0}\\ \\ \mathsf{0.475t^2 - 1.5t + 1.5=0}$

Now we solve for t:

$\mathsf{\Delta = (-1.5)^2 - 4\cdot0.475\cdot1.5}\\ \\ \mathsf{\Delta = 2.25-2.85=-0.6 \ \textless \ 0}$

We have a negative Delta. Therefore, there is no instant t when the dog reaches the cat.