Using the following equations as a guide (some may be used, some may not-choose wisely)

y1 OR x1= initial position
x2 or Y2 =final position
V1 or Vi= Initial velocity
V2 or Vf- Final velocity
a= Acceleration
T- time

Equation 1.) x2=x1 +vt

Equation 2.) v2= v1+at

Equation 3.) AvgV=(v1+v2)/2
*acceleration must be constant

Equation 4.) y2=y1 +V1t +1/2 at^2

Equation 5.) V2^2=v1^2 +2a (x2-x1)
*acceleration must be constant

1.) Two cyclists start out on the same trail together that is
overall 20km long. The first cyclist completes the entire
trail and is traveling at a constant 20 km/hr for the away
part of the trip and then only 15 km/hr for the return trip.
the Second cyclist, who is slower, is moving along at a
Constant 10 km/hr for the entire trip. The second cyclist
wants to turn around at some point so that he will arrive
back at the starting position at the same time that the
first cyclist will arive. At what point along the path
should he do that?

There do not seem to be any accelerations going on here.

First cyclist:
how long will he take"

outbound:
20 km / 20km/h = 1 h
inbound:
20 km / 15 km/hr = (4/3) h

total time for C1 = (7/3) hour

so
Second cyclist must do 2x km at 10 km/hr in (7/3) hr

2 x = 10 km/hr * (7/3) hr
x = 35/3 km
= 11 2/3 km out and the same back

To solve this problem, we can use Equation 1 and Equation 4.

Let's assume that the point where the second cyclist turns around is at distance x from the starting position.

For the first cyclist:
- The initial position (x1) is 0 km
- The final position (x2) is 20 km
- The initial velocity (v1) is 20 km/hr
- The final velocity (v2) is 15 km/hr

Using Equation 1, we can calculate the time taken by the first cyclist for the away part of the trip:
x2 = x1 + v1t
20 = 0 + (20)(t)
t = 1 hour

Now, using Equation 4, we can calculate the time taken by the first cyclist for the return trip:
x2 = x1 + V1t + (1/2)at^2
20 = 0 + (15)(t) + (1/2)(0)(t^2)
20 = 15t
t = 4/3 hours (approximately 1.33 hours)

Therefore, the total time taken by the first cyclist is 1 hour + 1.33 hours = 2.33 hours.

Now, let's find out when the second cyclist will arrive back at the starting position.

For the second cyclist:
- The initial position (y1) is 0 km
- The final position (y2) is also 0 km
- The initial velocity (v1) is 10 km/hr

Using Equation 4:
y2 = y1 + V1t + (1/2)at^2
0 = 0 + (10)(t) + (1/2)(0)(t^2)

Since the second cyclist travels at a constant speed for the entire trip, there is no acceleration (a = 0). Therefore, the equation simplifies to:
0 = 10t

This means that the second cyclist will arrive back at the starting position after t = 0 hours, which is immediately.

Therefore, the second cyclist should turn around and start heading back immediately to arrive back at the starting position at the same time as the first cyclist.