# Starting from a place, two person travel in bicycles along two perpendicular roads at speed of u km/hr and v km/hr. What is the distance between their positions after t hours?

## Does this problem use Pythagoras theorem? Do we also need to use Time-Distance-Speed formula? How do we connect this formula with Pythagoras theorem and solve the problem?

Oct 22, 2017

Use the definition of velocity (the time-distance-speed formula) to determine how far each went, then Pythagorus.

#### Explanation:

First, we calculate the distance each has travelled. Using $\Delta d = v \cdot \Delta t$

The rider travelling at $u$ km/hr has gone:

$\Delta {d}_{1} = u \cdot t$

The rider travelling at $v$ km/hr has gone:

$\Delta {d}_{2} = v \cdot t$

Since they are travelling along perpendicular directions, we must use Pythagorus' theorem to find the distance between them:

$\Delta {d}^{2} = \Delta {d}_{1}^{2} + \Delta {d}_{2}^{2} = {\left(u \cdot t\right)}^{2} + {\left(v \cdot t\right)}^{2}$

Finally, we take the square root of this equation, and simplify a bit:

$\Delta d = \sqrt{{\left(u \cdot t\right)}^{2} + {\left(v \cdot t\right)}^{2}}$

which we can also write as

$\Delta d = \sqrt{{u}^{2} + {v}^{2}} \cdot t$

Oct 22, 2017

$t \sqrt{{u}^{2} + {v}^{2}} k m s$

#### Explanation:

Let person 'x' starts from O and reached at A in t hours and at the same hours another person stars from O and reached Y.

First we have calculate distance between OA and OB. After that we have to calculate distance between AB.

We know, distance = speed x time.

For first person i.e. x speed is u km/hr and time is t hr.
so, distance (OA) = u km/hr x t hr = ut kms.

For second person, speed is v km/hr and time is t hr
So distance (OB) = v km/hr x t hr = vt kms.

Now, AOB has formed a right angle triangle whose height is OA and base is OB. We have to find the hypotenuse i.e. AB.

As per Pythagoras theory,
Hypotenuse = $\sqrt{h e i g h {t}^{2} + b a s {e}^{2}}$

Or, AB = $\sqrt{O {A}^{2} + O {B}^{2}}$

Or, AB = $\sqrt{{\left(u t\right)}^{2} + {\left(v t\right)}^{2}}$

Or, AB = $\sqrt{{u}^{2} {t}^{2} + {v}^{2} {t}^{2}}$

Or, AB = sqrt[t^2(u^2+v^2)

Or, AB = $t \sqrt{{u}^{2} + {v}^{2}}$

Hence the distance AB = $t \sqrt{{u}^{2} + {v}^{2}} k m s$