Two cyclists depart from the same point. One travels due East at 18km/hr while the other travels North-West at 20km/hr. How long will it be before they are 80km apart?
80^2=t^2(20^2+18^2)
solve for time t.
Wonderful, thank you
Two cyclists leave a park and ride in opposite directions, one averaging 5 mph and the other 10 mph. If they have two-way radios with a 10-mile range, for how many minutes will they remain in radio contact?
To find out how long it will take for the two cyclists to be 80km apart, we need to calculate the distance each cyclist travels over time.
Let's consider the two cyclists separately:
1. The cyclist traveling due East at 18km/hr.
Since the cyclist is traveling due East, their motion is only in the horizontal (x-axis) direction. We can calculate their distance using the formula: distance = rate × time.
Let's assume it takes this cyclist t hours to travel the distance of 80km. So, the equation for this cyclist will be: distance = rate × time, which gives us 80km = 18km/hr × t.
2. The cyclist traveling North-West at 20km/hr.
Since the cyclist is traveling North-West, their motion can be divided into two components: vertical (y-axis) and horizontal (x-axis). The speed in each direction will be the same, i.e., 20km/hr.
To find the distance traveled by this cyclist, we can use Pythagoras' theorem. The distance can be calculated as: distance = √(vertical distance^2 + horizontal distance^2).
If we assume the time taken by this cyclist to be t hours as well, then we have the equation: distance = √(20km/hr × t)^2 + (20km/hr × t)^2 = 80km.
Now we can solve these two equations to find the value of t:
For the cyclist traveling due East: 80km = 18km/hr × t,
For the cyclist traveling North-West: 80km = √(20km/hr × t)^2 + (20km/hr × t)^2.
By solving these equations simultaneously, we can determine the value of t.