## Problem Restatement A body starts falling from rest. The distance it covers in the **last second** of its motion equals the distance it covers in the **first three seconds**. **Question:** What is the total time of motion of the body? --- ## Solution Let: - \( g \) = acceleration due to gravity - \( T \) = total time of motion (unknown) - The body starts from rest (\( u = \)) ### Step 1: Distance covered in first 3 seconds From the equation: \[ S = ut + \frac{1}{2} g t^2 \] For the first 3 seconds (\( t = 3 \), \( u = \)): \[ S_1 = + \frac{1}{2} g (3)^2 = \frac{9}{2}g \] --- ### Step 2: Distance covered in the last second Distance in the last second: \[ S_2 = \text{Distance in } T \text{ seconds} - \text{Distance in } (T-1) \text{ seconds} \] \[ = \left( \frac{1}{2}g T^2 \right) - \left( \frac{1}{2}g (T-1)^2 \right) \] \[ = \frac{1}{2}g \left[ T^2 - (T-1)^2 \right] \] \[ = \frac{1}{2}g \left[ T^2 - (T^2 - 2T + 1) \right] \] \[ = \frac{1}{2}g (2T - 1) \] --- ### Step 3: Equate distances \[ \boxed{S_2 = S_1} \] \[ \frac{1}{2}g (2T - 1) = \frac{9}{2}g \] \[ 2T - 1 = 9 \] \[ 2T = 10 \] \[ T = 5 \] --- ## **Final Answer** \[ \boxed{5\ \text{seconds}} \] The total time of motion of the body is **5 seconds**.