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?

Question

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?

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Answer 1 · 2017-10-22T14:28:52

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

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 · 2017-10-22T14:44:54

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

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 = $t sqrt(u^2+v^2) kms$