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

Options

A. `(uv)/(x)`
B. `(u^2)/(x)`
C. `(v^2)/(x)`
D. `(u^2+v^2)/(x)`

Answer

A

Explanation

The diagram shows the dog's velocity just before and just after the moment in question, with a very small angle between them. Extending those arrows to the path of cat forms an isosceles triangle with height x and width v dt. (velocity of dog is always directed towards cat.)
Magnitude of acceleratio nof dog at required moment
`=(|vec_2-vec_1|)/(dt)`
Now `|vecv_1-vecv_2|=udtheta`
and `dtheta=(vdt)/(x)`
From equation (i),(ii) and (iii),
required magnitude of acceleration`=(uv)/(x)`