Don I. answered • 09/02/17

Tutor

4.9
(31)
Experienced teacher willing to help all students

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

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.