Question

The two ships leave simultaneously in direction right angles at each other.The speed of one of them exceeds other by 1KM /hr.The distance b/w ships after 1hour is 29km .Find their speeds.

Answer 1

See diagram.

Let faster ship cover distance x km
A is the faster ship, B is the slower ship, C are both their original positions

Therefore,
BC = √(AB²-AB²) (Pythagoras theorem)
= √(29²-x²)
= √(841-x²)km

Time taken for both = 1 hr
∴ Speed of A = x/1 = x km/hr
∴ Speed of B = √(841-x²)/1 = √(841-x²) km/hr

As per question,
Speed of A = Speed of B + 1
=> x = √(841-x²)+1
=> x-1 = √(841-x²)
=> (x-1)² = (√(841-x²))² (squaring both sides)
=> x²-2x+1 = 841-x²
=> 2x²-2x-840 = 0
=> x²-x-420 = 0
=> x²-21x+20x-420 = 0
=> x(x-21)+20(x-21) = 0
=> (x+20)(x-21) = 0

∴ x = -20 or 21
discarding -ve value
=> Speed of A (x) = 21 km/hr

=> Speed of B (√(841-x²)) = √(841-21²) = √(841-441) = √400 = 20 km/hr