Problem: Given a car starting from rest, accelerating at a rate "f" through a distance "S", continuing at constant speed for time "t", and then decelerating at a rate "f/2" to come to rest. If the total distance traversed is "15S", what is "S"?

Solution:

To solve this problem, we will use the following kinematic equations:

- v = u + at (equation 1)
- s = ut + 1/2 at^2 (equation 2)
- v^2 = u^2 + 2as (equation 3)

Where,

- v = final velocity
- u = initial velocity (zero in this case)
- a = acceleration
- t = time
- s = distance

Step 1: Finding the time taken for acceleration and deceleration

Let's first find the time taken for the car to accelerate and decelerate.

- During acceleration:
- final velocity = v1
- acceleration = f
- distance = S
- Using equation 3, we get:
- v1^2 = 2fS
- v1 = sqrt(2fS)
- Using equation 1, we get:
- v1 = at
- t = v1/a
- t = sqrt(2S/f)/f
- Therefore, time taken for acceleration = sqrt(2S/f)/f

- During deceleration:
- initial velocity = v1
- final velocity = 0
- acceleration = f/2
- distance = S
- Using equation 3, we get:
- v1^2 = 2f/2S
- v1 = sqrt(fS/2)
- Using equation 1, we get:
- v1 = at
- t = v1/(f/2)
- t = 2v1/f
- t = 2sqrt(Sf/2)/f
- Therefore, time taken for deceleration = 2sqrt(Sf/2)/f

Step 2: Finding the time taken for constant speed

The car travels at a constant speed for time "t". Therefore,

- distance covered during constant speed = vt
- total distance - distance covered during acceleration and deceleration = vt
- 15S - 2S = vt
- t = 13S/v

Step 3: Solving for "S"

Now that we have the time taken for acceleration, deceleration, and constant speed, we can solve for "S".

- Distance covered during acceleration = S
- Distance covered during deceleration = S
- Distance covered during constant speed = vt = 13S/v
- Total distance = 15S
- Therefore,
- 2S + 13S/v + 2S = 15S
- 13S/v = 1S
- v = 13

Substituting v = 13 in the equations for time taken for acceleration and deceleration, we get:

- Time