Question

From the same place at 7 a.m. A started walking in the north at a speed of 4 km/hr. after 1 hour, B started cycling in the east at a speed of 8 km/hr. at what time will they be at a distance of 20 km apart from each other?

Answer 1

Let t = travel time of bicycle

then (t + 1) = Walking time

Distance = Speed × Time

This is a Pythagoras problem

`a^2 + b^2 = c^2`

Where a = 8t, bicycle distance

b = 4(t + 1), walking distance

c = 20, distnace apart in t hrs from 8 a.m.

∴ `(8t)^2 + [4(t + 1)]^2 = (20)^2`

`64t^2 + 16(t^2 + 1 + 2t) = 400`

`64t^2 + 16t^2 + 32t + 16 = 400`

`80t^2 + 32t - 384 = 0`

`16(5t^2 + 2t - 24) = 0`

`5t^2 + 2t - 24 = 0`
`5t^2 + (12 - 10) t - 24 = 0`

`5t^2 + 12t - 10 t - 24 = 0`
`5t^2 - 10t + 12 t - 2 4 = 0`
`5t(t - 2) + 12(t - 2) = 0`
`(t -2)(5t + 12) = 0`
t = 2

and `t = -12/15`which is not possible

Positive solution t = 2 hrs.

At 10 a.m. they will be 20 km apart.