Question

A particle starts from rest with uniform acceleration. It travels a distance x in first 2 sec and a distance y in next 2 sec. Then y=3x or 4x or 2x or x?​

Answer 1

Answer:

2x is the correct answer .

Explanation:

because, the distance traveled in the first trip is x in 2 sec, then in other 2 seconds the same distance is travelled again. So, the distance traveled y is equal to 2x.

Answer 2

Given:

Initial velocity of particle, u=0

(Since it starts from rest)

It travels a distance x in first 2 sec and a distance y in next 2 sec

To Find:

Value of y in terms of x

Solution:

Let the uniform acceleration of given particle be a.

We know that,

• According to second equation of motion for constant acceleration

$\pink{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

where,

s is the displacement of the body

u is the initial velocity

a is acceleration

t is time taken

Now,

Given particle has travelled 'x' distance in first 2 sec

So, on applying second equation of motion, we get

$\longrightarrow\mathrm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\mathrm{x=0(2)+\dfrac{1}{2}a(2)^{2}}$

$\longrightarrow\mathrm{x=\dfrac{1}{\cancel{2}}a\times\cancel{4}}$

$\longrightarrow\mathrm{x=2a}$

$\longrightarrow\mathrm{2a=x}$

$\longrightarrow\mathrm{a=\dfrac{x}{2}}-------(1)$

Also, given particle has covered 'y' distance in next 2 sec or first 4 sec

So, on applying second equation of motion, we get

$\longrightarrow\mathrm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\mathrm{y=0(4)+\dfrac{1}{2}a(4)^{2}}$

$\longrightarrow\mathrm{y=\dfrac{1}{\cancel{2}}a\times\cancel{16}}$

$\longrightarrow\mathrm{y=8a}$

From (1), we get

$\longrightarrow\mathrm{y=\cancel{8}\bigg(\dfrac{x}{\cancel{2}}\bigg)}$

$\longrightarrow\mathrm{\green{y=4x}}$

Hence, y= 4x