
Don I. answered 09/02/17
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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.