two ships leave in direction right angle. the speed of one exceeds the other by 1km per hour.the distance between after one hour is 29 km. find speeds

Question

two ships leave simultaneously in direction right angle to each other. the speed of one ship exceeds the other by 1km per hour.the distance between the ships after one hour is 29 km. find their speeds

Answer 1

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

 

29 km is the hypotenuse.