Answer

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**Hint**It can be easily solved by using the equation of motion . Remember distance travelled by the next 10s will be greater as compared to first 10s, as the body is accelerating constantly (increasing speed at constant rate).

**Complete step by step solution**

The equation of motion that gives relation between time, acceleration and distance is

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

Where s is the distance travelled by object

$u$ is the initial velocity

$a$ is the acceleration of the object

$t$ is the time taken to travel

As in first 10s :

$u = 0$ (object starts from rest)

$t = $10s

Let $a$ be the constant acceleration and ${X_1}$ be the distance travelled in first 10s

So putting values we get

${X_1} = 0 + \dfrac{1}{2}a{(10)^2}$

Therefore ${X_1} = 50a$

Now in next 10s :

As we know after the first 10s there would be some velocity which would be initial velocity of next 10s.

To find it we will use equation of motion $v = u + at$

Where v is final velocity, u is initial velocity, a is acceleration, t is time

Putting values

$v = 0 + a(10)$

$v = 10a$

Now this is the final velocity after first 10s which will act as initial velocity for next 10s

So now

$u = 10a$

$t = 10$

$t$ is taken as 10 because we just want to find distance in those 10s (next 10s)

As acceleration is constant so it will remain $a$ and let ${X_2}$ be the distance travelled in these next 10s

Putting values in equation of motion:

${X_2} = 10a(10) + \dfrac{1}{2}a{(10)^2}$

${X_2} = 150a$

Now we have both the values of ${X_1}$ as well as ${X_2}$

$\dfrac{{{X_1}}}{{{X_2}}} = \dfrac{{50a}}{{150a}} = \dfrac{1}{3}$

Hence ${X_1} = \dfrac{1}{3}{X_2}$

**Option D is correct answer**

**Note**Remember as the body is constantly accelerated so it means that in any next interval of time it will cover more distance as compared to the first interval of time(provided time intervals are equal).

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