In 5th second the distance travelled by object is S=u+a(2×5-1) /2=u+9a/2In 4th second distance S=u+a(2×4-1) /2 = u+7a/2U+9a/2=u+7a/2+2a/2 =u+7a/2+a5th =4th+a

Relation for the Distance Travelled by an Object with Uniform Acceleration in the Interval between 4th and 5th Second

To determine the relation for the distance travelled by an object moving with a uniform acceleration in the interval between the 4th and 5th second, we need to consider the equations of motion and the concept of average velocity.

Equations of Motion:
1. Position (s) at time (t) is given by the equation: s = ut + (1/2)at^2
2. Velocity (v) at time (t) is given by the equation: v = u + at
3. Average velocity (v_avg) in a given time interval (t) is given by the equation: v_avg = (u + v)/2

Understanding the Interval between 4th and 5th Second:
The interval between the 4th and 5th second can be defined as the time duration from the beginning of the 4th second to the end of the 5th second. In other words, it is the time interval from t = 4 seconds to t = 5 seconds.

Calculating the Distance Travelled:
To calculate the distance travelled by the object during this interval, we can use the average velocity formula.

1. Calculate the initial velocity (u) at the beginning of the 4th second:
- If the object is at rest at t = 0, then u = 0.
- If the object has a known initial velocity, substitute the value for u.

2. Calculate the velocity (v) at the end of the 5th second:
- Use the equation v = u + at, where t = 5 seconds.
- Substitute the known values of u, a, and t to find v.

3. Calculate the average velocity (v_avg) during the interval:
- Use the equation v_avg = (u + v)/2.
- Substitute the known values of u and v to find v_avg.

4. Calculate the distance (s) travelled during the interval:
- Use the equation s = v_avg * t, where t is the duration of the interval.
- Substitute the known values of v_avg and t to find s.

Example:
Suppose an object has an initial velocity (u) of 2 m/s and a uniform acceleration (a) of 3 m/s^2. We want to find the distance travelled during the interval between the 4th and 5th second.

1. Calculate the initial velocity:
- u = 2 m/s.

2. Calculate the velocity at the end of the 5th second:
- v = u + at
= 2 + 3 * 5
= 17 m/s.

3. Calculate the average velocity during the interval:
- v_avg = (u + v)/2
= (2 + 17)/2
= 19/2 m/s.

4. Calculate the distance travelled during the interval:
- s = v_avg * t
= (19/2) * 1
= 19/2 m.

Therefore, the distance travelled by the object during the interval between the 4th and 5th second is 9.5 meters