To solve this problem, we can use the equations of motion to find the ratio of the initial velocities of Akshat and Hariom.

Let's assume the initial velocity of Akshat is v1 and the initial velocity of Hariom is v2. The distance they need to travel is 1000m.

We are given that the ratio of their accelerations is 20:3. Let's assume the acceleration of Akshat is a1 and the acceleration of Hariom is a2.

1. Finding the time taken by Akshat:
Using the equation of motion: s = ut + 0.5at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.

For Akshat:
s = 1000m, u = v1, a = a1
1000 = v1 * t1 + 0.5 * a1 * t1^2 ----(1)

2. Finding the time taken by Hariom:
Since Hariom starts moving after tes (time taken by Akshat), his time taken will be less. Let's assume the time taken by Hariom is t2.

For Hariom:
s = 1000m, u = v2, a = a2
1000 = v2 * t2 + 0.5 * a2 * t2^2 ----(2)

3. Finding the ratio of initial velocities:
We need to find the ratio v1:v2.

Dividing equation (1) by equation (2), we get:
(v1 * t1 + 0.5 * a1 * t1^2) / (v2 * t2 + 0.5 * a2 * t2^2) = 1

4. Substituting the values:
We are given that the ratio of accelerations is 20:3, so we can substitute a1 = 20x and a2 = 3x, where x is a constant.

Let's assume t1 = t and t2 = t - tes, where tes is the time taken by Akshat.

Substituting these values in equation (1) and (2), we get:
1000 = v1 * t + 0.5 * 20x * t^2 ----(3)
1000 = v2 * (t - tes) + 0.5 * 3x * (t - tes)^2 ----(4)

5. Solving equations (3) and (4):
We have two equations (3) and (4) with two unknowns v1 and v2. We can solve these equations simultaneously to find the ratio v1:v2.

By solving equations (3) and (4), we can find the values of v1 and v2. Then, we can divide v1 by v2 to get the ratio of initial velocities.