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Two ports A and B are 300 km apart. Two leave A for B such that the 2nd leaves 8 hours after the first. The ships arrive at B simultaneously.
Find the time the ship spent on the trip if the speed of one of them is 10 km/h higher than that of the other.
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Let u = the speed of the slower ship.
Then the speed of the faster ship is u + 10.
The slower ship spent hours for the trip.
The faster ship spent hours.
According to the condition, the difference of these times is 8 hours. It gives an equation
(((300/u}}} - = 8.
To solve it, multiply both sides by u*(u+1) and then simplify.
You will get
300*(u+10) - 300u = 8u*(u+10),
3000 = + ,
+ - = .
+ - = .
Factorize:
(u-15)*(u+25) = 0.
The roots are u = 15 and u = -25.
The solution is 15 for the slower ship.
The time is = 20 hours for the slower ship.