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#1 2017-01-18 18:44:25

LC
Guest

My son's problem

My son's math class had them working on the problem with groups of four students.  He just informed that he has been working on it as the only member of his group.  As I tried to help him I was lost past the basic math.


The problem:
People leaving for a cruise got to dock late, the ship had already travelled five hours.  Ship was 100 miles away when they got to dock. A speedboat awaits at dock which can travel 200 miles in five hours.  A helicopter is available which travels at 90 miles and hours, but they have to wait 3.5 hours before being picked up.

What should they do to catch up with the ship?
My son was instructed to make tables and graphs to model the situation.

Any suggestions?

Thanks

#2 2017-01-18 21:06:31

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

Re: My son's problem

hi LC

The graph he needs is called a distance / time graph.

Mark hours on the across axis, 0 to 15 should do it.

On the up axis put distance from the dock, 0  to 250 should be enough.

Start the ship at (0,0) and also mark a point at (5,100) to show the ship's position when the latecomers arrive at the dock.

Join these points and continue it as the ship will carry on sailing away.

The latecomers are at (5,0) as they are at the dock, but 5 hours after the ship sailed.

If they use the speedboat they will travel 200 miles in 5 hours.  That will get them to the point (10, 200) so mark that point and join up their line of travel.  Where the two lines cross is the catch up moment and place.

If they wait 3.5 hours that will make the time 8.5 hours so mark a point at (8.5, 0).  One hour later the helicopter can be at (9.5,90) so mark that point, Draw the line of travel for the helicopter and extend it until it cuts the ship's line.  That will give the alternative catchup point.

So you can then decide which to go for.  Hidden is my attempt. The helicopter line is not quite accurately done but it'll give you an approximation.  Get him to try it first and then you can check it yourself. 

The graph should be enough to make the decision.  The catchup time for the speedboat option is easy to calculate.

For the helicopter it will require some algebra. Use the formula distance = speed x time   Call the meetup time T.  The ship has travelled 20T as it is going at 20 mph.  The helicopter starts at 8.5 and ends at T so its time of travel is (T-8.5).  It is going at 90 mph so it will travel 90.(T-8.5)

If you put these two distances equal (we want the helicopter to be at the same spot as the ship) and solve for T you will have the catchup time for the helicopter.

I would do both these calculations as it 'proves' by calculation which is the better option, and avoids any graphical inaccuracies.

Bob

ps.  I'm an ex-teacher, so naturally, I want the rest of his group to do the work too.  smile  If you think he has got it right, he could sit them down and instruct them in how to solve the problem.  This is a good way to test if you have understood something.  If you can teach others to do it, then you must have got the idea and understood it well.  If they won't cooperate he should submit his answer and tell the teacher the others haven't contributed.  Then it's up to the teacher to deal with them.  They may suffer a short term annoyance but in the end will respect him more (I hope; but if not, then they weren't very good friends any way)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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