Cyclist and a regular bus service.

let the speed of bus be x, then the distance between buses "covered" in 6 minutes with the speed of x+20 and we need 18 minutes for the same distance with speed x - 20

therefore x + 20 = 3 * (x - 20) or x = 40 (km/h)

since we cover the distance between buses in 18 minutes with speed 20 km/h, we need 9 minutes to cover the distance with the speed of bus, so buses leave each 9 minutes

I used $T_{f}$ to represent the period of buses coming towards him from his 'front,' and $T_{b}$ to represent the period of buses coming towards him from 'behind.'

We can essentially treat the buses as "wave crests" of extremely long sound waves, which means we can apply doppler-effect formulas.

The frequency doppler effect formula for an observer moving towards a stationary source is given as:

$f_{observed} = f_{source}\frac{v_{sound}+v_{observer}}{v_{sound}}$

Instead of $v_{sound}$ , I used $v_{B}$ for the speed of the bus. $v_{observer}$ is simply the speed of the cyclist, or $v_{c}$ . Frequency is the inverse of period; that is, $f_{observed} = \frac{1}{T_{f}}$ , and $f_{source} = \frac{1}{T}$ . This gives us:

$\frac{1}{T_{f}} = \frac{1}{T}\frac{v_{B}+v_{c}}{v_{B}}$

The doppler effect formula for an observer moving away from a stationary source is:

$f_{observed} = f_{source}\frac{v_{sound}-v_{observer}}{v_{sound}}$

Again, making the appropriate substitutions, this becomes:

$\frac{1}{T_{b}} = \frac{1}{T}\frac{v_{B}-v_{c}}{v_{B}}$

I took each equation separately and solved for $v_{B}$ , which game me:

$v_{B} = \frac{T_{f}v_{c}}{T-T_{f}}$ , and $v_{B} = \frac{T_{b}v_{c}}{T_{b}-T}$

Setting these equations equal to eachother (interestingly) cancels out $v_{c}$ . We get:

$\frac{T_{f}}{T-T_{f}} = \frac{T_{b}}{T_{b}-T}$

Solving this for $T$ we get:

$T = \frac{2T_{f}T_{b}}{T_{f}+T_{b}} = \frac{2(6min)(18min)}{24min} = \boxed{T = 9min}$

This problem is flawed. There are two possible solutions: 18 or 9.

Both buses need to have the same speed on the road. The conditions of the problem are satisfied if both buses travel at 10 mph or if both buses travel at 40mph. In the first case, the oncoming buses travel at 30mph backwards from his reference frame, and the buses in his direction travel at 10mph backwards. The oncoming buses travel at 3x the speed of the other buses, so the conditions are satisfied. From repeated substitutions into $d=rt$ , we find that there are 3 miles between buses, and 18 minutes between departures. However, the biker is faster than the buses in this case, which is probably not very likely.

In the second case, the oncoming buses travel at 60mph backwards from the biker's perspective, and the buses in his direction travel at 20mph forwards. Again, the problem's conditions are satisfied. We find that there are 6 miles between buses and 9 minutes between launches, which is the "correct" answer to this problem. However, the problem should be modified to say that the buses have a higher absolute speed than the biker, or so that both solutions, 9 and 18, are accepted.

The problem does state that the slower buses go past 'in the direction of him', but this should be clarified to mean 'from his perspective', as it could ambiguously mean 'relative to the road'.

Let's find the speed of the buses. Let's assume it to be x for now. When the rider completes 18 min. of his journey,a bus overtakes him. Relative velocity = x-20 ( rider's speed = 20km/h) But distance covered by both bus and rider would same which is 6km. [Solving for distance using rider's velocity 20km/h and t=3/10 hr (18min.)] Thus, x-20=6/3/10 (Speed = Distance/Time) x-20=20 x=40km/hr Hence, Time = Distance / Speed Time=6/40 ×60 =360/40= 9 min.

b=bus speed t=time

bt=6(b+20)=18(b-20) bt =6b+120=18b-360 12b=480,b=40,t=360/40=9

bus speed = 40 km/h time = 9 min