80^2=t^2(20^2+18^2)
solve for time t.
solve for time t.
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.