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(A) ${y_2} = 2{y_1}$

(B) ${y_2} = 3{y_1}$

(C) ${y_2} = 4{y_1}$

(D) ${y_2} = 5{y_1}$

Answer

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Useful formula:

The acceleration equation of the motion is given by,

$s = ut + \dfrac{1}{2}a{t^2}$

Where, $s$ is the distance travelled by the particle, $u$ is the initial velocity of the particle, $t$ is the time taken by the particle to travel the distance and $a$ is the acceleration of the particle.

The velocity of the particle is given by,

$v = at$

Where, $v$ is the final velocity of the particle, $a$ is the acceleration of the particle and $t$ is the time taken by the particle.

Given that,

The particle starts from rest that means, $u = 0\,m{s^{ - 1}}$,

The particle moves in the constant acceleration is, $a$,

The time taken for the ${y_1}$ distance is, $t = 10\,\sec $,

The time taken for the ${y_2}$ distance is, $t = 10\,\sec $.

Now,

The acceleration equation of the motion for the ${y_1}$ distance is given by,

${y_1} = ut + \dfrac{1}{2}a{t^2}\,.....................\left( 1 \right)$

By substituting the initial velocity, time taken for the distance ${y_1}$ and the acceleration in the above equation (1), then the above equation is written as,

${y_1} = \left( {0 \times 10} \right) + \dfrac{1}{2}a \times {10^2}$

By multiplying the terms in the above equation, then

${y_1} = 0 + \dfrac{1}{2}a \times {10^2}$

By squaring the terms in the above equation, then

${y_1} = \dfrac{1}{2} \times 100a$

By dividing the terms in the above equation, then

${y_1} = 50a$

Now, the velocity of the particle is given by,

$v = at\,................\left( 2 \right)$

By substituting the acceleration and the time in the above equation, then

$v = 10a$

By substituting this velocity in the equation (1), then the distance will become ${y_2}$, then

${y_2} = \left( {10a \times 10} \right) + \dfrac{1}{2}a \times {10^2}$

By multiplying the terms in the above equation, then

\[{y_2} = 100a + \dfrac{1}{2}a \times {10^2}\]

By squaring the terms in the above equation, then

\[{y_2} = 100a + \dfrac{1}{2} \times 100a\]

By dividing the terms in the above equation, then

\[{y_2} = 100a + 50a\]

By adding the terms in the above equation, then

\[{y_2} = 150a\]

The above equation is also written as,

\[{y_2} = 3 \times 50a\]

By substituting the equation of the ${y_1}$ in the above equation, then

\[{y_2} = 3{y_1}\]