Question

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?

Answer 1

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 `Deltad=v*Deltat`

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

`Deltad_1=u*t`

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

`Deltad_2=v*t`

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

`Deltad^2=Deltad_1^2+Deltad_2^2=(u*t)^2+(v*t)^2`

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

`Deltad=sqrt((u*t)^2+(v*t)^2)`

which we can also write as

`Deltad=sqrt(u^2+v^2)*t`

Answer 2

`t sqrt(u^2+v^2) kms`

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 (height^2 + base^2)`

Or, AB = `sqrt(OA^2+OB^2)`

Or, AB = `sqrt[(ut)^2+(vt)^2]`

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