Question

A cat runs along a straight line with a constant velocity of magnitude $\mathrm{v}$. A dog chases the cat such that the velocity of the dog is always directed towards the cat. The speed of a dog is $\mathrm{u}$ and always constant. At the instant, both are separated by distance $\mathrm{x}$ and their velocities are mutually perpendicular, the magnitude of the acceleration of the dog is.

• A. $\frac{{\mathrm{u}}^2}{\mathrm{x}}$
• B. $\frac{\mathrm{uv}}{\mathrm{x}}$
• C. $\frac{{\mathrm{v}}^2}{\mathrm{x}}$
• D. $\frac{\left({\mathrm{u}}^2+{\mathrm{v}}^2\right)}{\mathrm{x}}$

Answer 1

The correct option is B

$\frac{\mathrm{uv}}{\mathrm{x}}$

Step 1. Given data

Velocity of cat $=\mathrm{v}$

Speed of dog $=\mathrm{u}$

Step 2: Formula used

Acceleration is defined as the rate of change of velocity, hence mathematically in terms of initial and final velocity it can be written as,

$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{{\mathrm{u}}_2}-\overrightarrow{{\mathrm{u}}_1}}{\mathrm{dt}}$

Step 2. Find the magnitude of the acceleration of the dog.

The above diagram represents the dog's velocity just before and just after the moment given, with a very small angle between them. Extending those arrows to the path of the cat creates an isosceles triangle with height $\mathrm{x}$ and width $\mathrm{v}\times \mathrm{dt}$(the velocity of the dog is always directed towards the cat.) The magnitude of the acceleration of the dog at the required moment.

Let the velocity of the dog be ${\mathrm{u}}_1$, and after some time its velocity is ${\mathrm{u}}_2$ which is shown in the above figure.

By using vector properties,

$\mathrm{vd\theta }={\mathrm{u}}_2-{\mathrm{u}}_1$ ------ $\left(1\right)$

So,

$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{{\mathrm{u}}_2}-\overrightarrow{{\mathrm{u}}_1}}{\mathrm{dt}}$

From the equation $\left(1\right)$ we get,

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{vd\theta }}{\mathrm{dt}}$

When $\mathrm{d\theta }$ is very small, then

$\mathrm{d\theta }=\frac{\mathrm{arc}\mathrm{length}}{\mathrm{radius}}$

$\mathrm{d\theta }=\frac{\mathrm{v}\mathrm{dt}}{\mathrm{x}}$

$\mathrm{a}=\frac{\mathrm{v}\mathrm{dt}}{\mathrm{x}\mathrm{dt}}$

$=\frac{\mathrm{uv}}{\mathrm{x}}$

So, the magnitude of the acceleration of the dog is $\frac{\mathrm{uv}}{\mathrm{x}}$.

Hence, option $\mathrm{B}$ is the correct answer.