Understanding Uniform Acceleration and Distances Travelled

This problem asks us to find a relationship between two distances, $s_1$ and $s_2$, covered by a body under uniform acceleration. The body starts from rest, which is a crucial piece of information. We need to use the laws of motion for uniformly accelerated objects.

Key Concepts Used:

• Equations of Motion: For an object moving with uniform acceleration, the distance ($s$) covered in time ($t$) is given by the formula: $ s = ut + \frac{1}{2}at^2 $ where $u$ is the initial velocity and $a$ is the uniform acceleration.
• Starting from Rest: This means the initial velocity ($u$) is 0.
• Distances in Specific Time Intervals: We need to calculate the distance covered in the first 2 seconds and the distance covered in the subsequent 4 seconds.

Calculating Distance $s_1$

The problem states that the body starts from rest, so $u = 0$. We are given that the distance travelled in the first 2 seconds is $s_1$. Let's use the equation of motion:

Time, $t = 2$ seconds.

Using the formula $s = ut + \frac{1}{2}at^2$:

$ s_1 = (0)(2) + \frac{1}{2}a(2)^2 $ $ s_1 = 0 + \frac{1}{2}a(4) $ $ s_1 = 2a $

So, the distance covered in the first 2 seconds is $s_1 = 2a$. We can express acceleration in terms of $s_1$ as $a = \frac{s_1}{2}$.

Calculating Distance $s_2$

The distance $s_2$ is covered in the *next* 4 seconds. This means the time interval for $s_2$ is from $t=2$ seconds to $t=6$ seconds.

To find $s_2$, we first calculate the total distance travelled in the first $2 + 4 = 6$ seconds. Let's call this total distance $s_{total}$.

For the total time $t = 6$ seconds:

$ s_{total} = ut + \frac{1}{2}at^2 $ $ s_{total} = (0)(6) + \frac{1}{2}a(6)^2 $ $ s_{total} = 0 + \frac{1}{2}a(36) $ $ s_{total} = 18a $

Now, the distance $s_2$ (covered in the next 4 seconds) is the difference between the total distance covered in 6 seconds ($s_{total}$) and the distance covered in the first 2 seconds ($s_1$):

$ s_2 = s_{total} - s_1 $ $ s_2 = 18a - 2a $ $ s_2 = 16a $

Finding the Relation Between $s_1$ and $s_2$

We have derived the following expressions:

• $s_1 = 2a$
• $s_2 = 16a$

We need to find a direct relation between $s_1$ and $s_2$. We can substitute the value of $a$ from the first equation into the second equation.

From $s_1 = 2a$, we get $a = \frac{s_1}{2}$.

Now substitute this into the equation for $s_2$:

$ s_2 = 16 \left( \frac{s_1}{2} \right) $ $ s_2 = \frac{16}{2} s_1 $ $ s_2 = 8s_1 $

Conclusion

The relation between the distance $s_1$ travelled in the first 2 seconds and the distance $s_2$ travelled in the next 4 seconds is $s_2 = 8s_1$.