Just plug in your distance formula and solve for t when d=50. At t hours past noon,
d = √((10t)^2 + (20(t-1))^2)
d = 10√(5t^2-8t+4)
when d=50,
5t^2-8t+4 = 25
5t^2-8t-21 = 0
(5t+7)(t-3) = 0
we want t>0, so at t=3 (3:00 pm) the ships are 50 miles apart
Note that the negative value of t means that if the ships had been sailing on the same courses, rather than just starting out, they would also have been 50 miles apart at t=-7/5, or at 10:36 am. Of course, that might have been awkward, coming at the port from the land side!
A ship leaves a port at 12:00 Noon and sails East at speed of 10 miles/hour. Another ship leaves
the same port at 1:00 PM and sails North at a speed of 20 miles/hour. At what time are the two
ships going to be 50 miles apart from each other? (Hint: Distance = Speed * Time)
3 answers
I did not anderstand the answer
the paths of the ships form two legs of a right triangle, where the distance between the ships is the hypotenuse. As usual, draw a diagram.
Now, the hypotenuse of a triangle with legs a and b is
d = √(a^2+b^2)
The length of the legs is the distance traveled by the ships, which is each ship's speed times the # hours spent sailing.
The 10 km/hr ship has been going for t hours, but the other ship waited an hour before starting out, so its sailing time is only t-1 hours.
Now, the hypotenuse of a triangle with legs a and b is
d = √(a^2+b^2)
The length of the legs is the distance traveled by the ships, which is each ship's speed times the # hours spent sailing.
The 10 km/hr ship has been going for t hours, but the other ship waited an hour before starting out, so its sailing time is only t-1 hours.
0 answers