Problem Statement:
A man is standing 40m behind a bus. The bus starts with a constant acceleration of 1m/s² and at the same instant, the man starts moving with a constant speed of 9m/s. We need to find the time taken by the man to catch the bus.

Analysis:
To solve this problem, we need to find the time it takes for the man to cover the initial distance of 40m between him and the bus, considering their respective speeds and accelerations.

Solution:

Step 1:
Let's assume that the man catches the bus after time 't'. We need to find this value of 't'.

Step 2:
We know that the man starts with an initial velocity of 0 and moves with a constant speed of 9m/s. Therefore, the distance covered by the man in time 't' is given by:
Distance covered by the man = Speed × Time
Distance covered by the man = 9t

Step 3:
The bus starts with an initial velocity of 0 and has a constant acceleration of 1m/s². Therefore, the distance covered by the bus in time 't' is given by:
Distance covered by the bus = Initial Velocity × Time + (1/2) × Acceleration × Time²
Distance covered by the bus = 0 × t + (1/2) × 1 × t²
Distance covered by the bus = (1/2) × t²

Step 4:
Since the man catches the bus, the distance covered by the man should be equal to the distance covered by the bus. Therefore, we equate the two distances calculated in Step 2 and Step 3:
9t = (1/2) × t²

Step 5:
To solve the equation, we bring all the terms to one side:
(1/2) × t² - 9t = 0

Step 6:
Simplifying the equation further, we get:
t² - 18t = 0

Step 7:
Factorizing the equation, we get:
t(t - 18) = 0

Step 8:
From the equation, we have two possible solutions:
t = 0 (which is not possible in this context)
t - 18 = 0
t = 18

Step 9:
Therefore, the time taken by the man to catch the bus is 18 seconds.