A body starts from rest, the ratio of distances travelled by the body during 3rd and 4th seconds is -

To find the ratio of distances traveled by a body during the 3rd and 4th seconds, we need to understand the concept of motion and acceleration.

Assuming the body starts from rest, we can use the equations of kinematics to find the distance traveled during a specific time interval.

Let's denote the distance traveled by the body during the 3rd second as d3 and during the 4th second as d4.

Since the body starts from rest, we know that the initial velocity (u) is 0.

The equation that relates the distance traveled (d), initial velocity (u), time (t), and acceleration (a) is:

d = ut + (1/2)at^2

In this case, the acceleration can be calculated by the formula:

a = (v - u) / t

where v represents the final velocity. Since the body starts from rest, the final velocity v would also be 0.

Now, let's calculate the distances traveled during the 3rd and 4th seconds:

For the 3rd second:
t = 3s
u = 0
v = 0

a = (v - u) / t
= (0 - 0) / 3
= 0

Using the equation d = ut + (1/2)at^2:
d3 = 0 * 3 + (1/2) * 0 * (3^2)
= 0

So, the distance traveled during the 3rd second is 0.

For the 4th second:
t = 4s
u = 0
v = 0

a = (v - u) / t
= (0 - 0) / 4
= 0

Using the equation d = ut + (1/2)at^2:
d4 = 0 * 4 + (1/2) * 0 * (4^2)
= 0

So, the distance traveled during the 4th second is also 0.

Therefore, the ratio of distances traveled by the body during the 3rd and 4th seconds is 0:0, which simplifies to 0:0 or 0.

if so then

v = -gt
d = -(1/2) g t^2

from second two to second three:
d = -(1/2)g (9-4) = -(5/2) g

from second three to second four:
d = -(1/2)g (16-9) = -(7/2) g

ratio = 7/5