Answer to Question #138515 in Physics for Annie

Question

A dog is at a location 12.1 m behind its owner (x= -12.1m). The owner throws the ball 3m ahead of themselves for their dog to chase (x=3m). The dog is able to run with a velocity of 5 m/s. (A) Where is the dog after 1.6 s of running? (B) How long does it take for the dog to reach the ball?

Answer 1

The position of the dog is given by the following:

$$x= x_0 + vt$$

where $x_0 = -12.1 m$ is the intial position and $v = 5m/s$. Thus, after $t = 1.6s$ the dog will be in:

$$x = -12.1m + 5m/s\cdot 1.6s = -4.1m$$

The distance between the dog and the ball is:

$$d = 3m-(-12.1m) = 15.1m$$

Moving with the velocity $v = 5m/s$ the dog covers this distance in:

$$t = \dfrac{d}{v} = \dfrac{15.1m}{5m/s} =3.02s$$

Answer. a) -4.1m, b) 3.02 s.