Question

A man leaves a point walking at 6.5 km/h in a direction E 20◦ N (i.e. a bearing of 70◦ ). A cyclist leaves the same point at the same time in a direction E 40◦ S (i.e. a bearing of 130◦ ) travelling at a constant speed. Find the average speed of the cyclist if the walker and cyclist are 80 km apart after 5 hours​

Answer 1

Answer:

Step-by-step explanation:

Let's rotate our reference frame 40° CW so that the cyclist is traveling along the x axis and the walker is traveling along θ = Ε 60° Ν

Let "v" be the average cyclist speed

The rate of change of north position is 6.5sin60 m/s

In five hours, the walker has traveled north 5(6.5sin60) = 32.5sin60 km

The relative east velocity is v - 6.5cos60 = v - 3.25

In five hours, the cyclist has changed east position relative to the walker by 5(v - 3.25) = (5v - 16.25) km.

80² = (32.5sin60)² + (5v - 16.25)²

6400 = 792.1875 + 25v² - 162.5v + 264.0625

0 = 25v²- 162.5v - 5343.75

quadratic formula, positive answer

v = (162.5 + √(162.5² - 4(25)(-5343.75))) / (2(25))

v = (162.5 + 748.85329) / 50

v = 18.227065...

v = 18.2 km/h