Solution to Question 39
Concept
The dog runs with a constant speed $v$ and always heads towards the cat. This implies that the dog’s velocity vector is always aligned with the position vector $\vec{r}$ pointing from the dog to the cat.
We need to find the acceleration of the dog. Since the dog’s speed $v$ is constant, it has no tangential acceleration ($a_t = \frac{dv}{dt} = 0$). The only acceleration is centripetal (or normal) acceleration due to the change in direction of motion.
Calculation
The direction of the dog’s motion is the line of sight. Therefore, the rate at which the dog turns is equal to the angular velocity $\omega$ of the line of sight (the line connecting the dog and the cat).
We calculate $\omega$ by considering the components of velocity perpendicular to the line of sight:
- The dog’s velocity component perpendicular to the line is $0$ (since it runs along the line).
- The cat’s velocity is $u$. Its component perpendicular to the line of sight is $u \sin\theta$.
The angular velocity of the rotating line of sight is:
$$ \omega = \frac{v_{perp}}{r} = \frac{u \sin\theta}{r} $$The dog’s normal acceleration is given by $a_n = v \omega$.
$$ a = v \left( \frac{u \sin\theta}{r} \right) = \frac{uv \sin\theta}{r} $$