A cat is sleeping on the floor in the middle of a 2.8-m-wide room when a barking dog enters with a speed of 1.40 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/s^2 as soon as it enters the room. Required:
How far is the cat in front of the dog as it leaps through the window?

the cat is  0.4238 m in front of the dog as it leaps through the window

Explanation:

Given that;

acceleration a = 0.85 m/s²

speed v = 1.40 m/s

the cat is at rest, so initial velocity u = 0  

we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;

using the second equation equation of motion;

s = ut + 1/2 at²  

we substitute

2.8/2 = 0×t + 1/2 × 0.85 × t²

1.4 = 0.425t²

t = √( 1.4 / 0.425 )

t = 1.81497 sec

now, at acceleration 0.10 m/s²

the dog has to cover the distance;

s = ut + 1/2 at²  

s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²

s =  2.540958 - 0.1647

s = 2.3762 m  

The cant in front of the dog as it leaps through the window;

distance = 2.8 m - 2.3762 m

distance = 0.4238 m

Therefore, the cat is  0.4238 m in front of the dog as it leaps through the window

heat the gas and measure the pressure

2b2t

explanation:

2b2t