Question

A body starts moving with uniform acceleration from its state of rest. If it covers distance x in first 2 second and distance y in next 2 second, obtain the relation between x and y.

Answer 1

EXPLANATION:-

u= 0 m/s

[KINDLY REFER TO THE ATTACHMENT FOR PATH OF THE OBJECT ]

Using 2nd equation:-

$\sf\;x=ut+\frac{1}{2}at^2$

$\sf\;x=0\times2+\frac{1}{2}a(2)^2\\\\\boxed{\text{x=2a}}$

Y can't be calculated as initial velocity is not known .So we will first calculate AB point acceleration and then subtract AC point acceleration .

$\sf\;AB=ut+\frac{1}{2}at^2\\\\AB=\frac{1}{2}\times16a\\\\AB=8a$

Now ,

CB point acceleration=AB acceleration-AC acceleration

CB point acceleration=8a-2a

CB point acceleration=6a

So :-

$\sf\;\frac{x}{y}=\frac{2a}{6a}\\\\\frac{x}{y}=\frac{1}{3}\\\\3x=y$

Hence the relation between x and y is that 3 times of x is y or 3x=y .