Question

Four particles are located at the vertices of a
square of side "a'. They all start moving
simultaneously, at t = 0, with the same speed
V, with first particle heading for second,
second heading for third and so on.
Then the acceleration of each particle at
t=a/4v is​

Answer 1

Answer:

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Answer 2

Given:

sides of square = a

To find :

The acceleration of each particle at t =?

Explanation:

At t = 0, first particle heading for a second,  second heading for third and so on with the same speed V.

∴ ΔV = $\frac{V}{3}$              ...(1)

We have to calculate acceleration

∴  $\displaystyle a =\frac{\triangle V}{t}$

from equation(1)

$\displaystyle a =\frac{\frac{V}{3} }{\frac{a }{4V} }$

$\displaystyle a = \frac{V}{3} \times \frac{4V }{a}$

$\displaystyle \mathbf{ a = \frac{4V^2}{3a} }$