Question

A particle starts from rest with constant acceleration for $20\,\sec $. If it travels a distance ${y_1}$ in the first $10\,\sec $ and a distance ${y_2}$ in the next $10\,\sec $, then?
(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 1

Hint: The relation between the ${y_1}$ and the ${y_2}$ can be determined by using the acceleration equation of the motion. The initial velocity is taken as zero because the particle starts from rest, so the initial velocity is taken as zero, then the relation between the ${y_1}$ and the ${y_2}$ can be determined.
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.

Complete step by step solution:
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
\[{y_2} = 3{y_1}\]

Hence, the option (B) is the correct answer.

Note: The velocity of the distance ${y_2}$ is not zero because, the distance ${y_2}$ is started when the particle is moving after the distance of the ${y_1}$, so that there must be the velocity of the particle exists. But in the distance ${y_1}$ the velocity is zero because it is the starting point.