Question

A sitting cat in a field suddenly sees a standing dog. To save its life the cat runs away in a straight line with constant speed $u$. Without any time lag, the dog starts with a constant speed $v$ to catch the cat. At the initial moment, $v \bot u$ and $l$ is the separation between them. If the dog always heads towards the cat, find the time after which the dog catches the cat.

Answer 1

In this solution, we will use the concepts of relative velocity. To catch up with the cat, the dog must travel a horizontal distance between it and the cat. The relative velocity of the dog with respect to the cat should be such that it covers the distance between them in the same amount of time too.

Complete step by step answer:
We’ve been given that a cat sees a dog and starts running perpendicular to the distance between them. Then, the dog starts running after the cat such that initially, it is in the direction of the cat but then since it keeps on heading towards the cat, its direction changes as shown below.

Now the vertical component of the dog’s velocity will be ${v_y} = v\sin \theta$
The dog must cover a vertical distance equal to the distance travelled by the cat to catch up to the cat. Let the time in which this happens to be denoted by $t$. Then the relation between distance and velocity for the dog can be written as
$d = \int {v\sin \theta .dt}$
In the same amount of time, the cat will travel a distance of $u \times t$ since it is already going vertically.
Then we can write
$\int {v\sin \theta .dt} = ut$
Since the speed of the dog is constant with time,
$\int {\sin \theta .dt} = \dfrac{{ut}}{v}$
Now, we can also say that to catch the cat, the relative velocity of the dog with respect to the cat in the vertical direction should be such that it catches up to the cat in time $t$ . Since the relative velocity of the cat with respect to the dog is $v - u\sin \theta$, we can write
$\int {(v - u\sin \theta )} dt = l$
This integral can be broken down into
$\int {vdt} - u\int {\sin \theta dt} = l$
As $\int {\sin \theta .dt} = \dfrac{{ut}}{v}$, we can write
$vt - u\dfrac{{ut}}{v} = l$
Solving for $t$, we get
$t = \dfrac{{Lv}}{{{v^2} - {u^2}}}$

Note: We should be familiar with the concepts of relative velocity to solve such questions. While we don’t need to know the entire trajectory of the dog and the cat, we need to have a general idea of how the trajectory will evolve to determine the relations between their velocities as we did above.