## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the time at which the two animals have the same speed. $(0.85~m/s^2)~t = (1.5~m/s)-(0.10~m/s^2)~t$ $t = \frac{1.5~m/s}{0.95~m/s^2}$ $t = 1.58~s$ We can find the distance the cat moves in this time. $x_c = \frac{1}{2}at^2$ $x_c = \frac{1}{2}(0.85~m/s^2)(1.58~s)^2$ $x_c = 1.06~m$ We can find the distance the dog moves in this time. $x_d = v_0t+\frac{1}{2}at^2$ $x_d = (1.5~m/s)(1.58~s)-\frac{1}{2}(0.10~m/s^2)(1.58~s)^2$ $x_d = 2.25~m$ At t = 1.58 s, the cat is 0.44 meters from the window and the dog is 0.75 meters from the window. After this time, the cat is faster than the dog so the cat can escape out the window.