How long does it take mercutio to reach the y axis

Question

Juliet and Mercutio are moving at constant speeds in the xy-plane. They start moving at the same time. Juliet starts at the point (0, −6) and heads in a straight line toward the point (10, 5), reaching it in 10 seconds. Mercutio starts at (7, −18)
and moves in a straight line. Mercutio passes through the same point on the x-axis as Juliet, but 2 seconds after she does.

how long does it take mercutio to reach the y axis?

 

Answer 1

Hi Jon,

 

There's quite a bit to unpack in this problem, isn't there?  I can see why it's giving you a hard time. 

 

Well now, here's the plan: we know how fast Juliet is moving, because we know how long it takes her, and we can figure out how far she goes. Juliet and Mercutio both pass through some point (0, x) on the x axis.  We can find this point and figure out what time tJ Juliet moved through (0, x), since we can calculate how far it is from her starting point, and we know how fast she is moving.  We know that Mercutio moves through (0, x) at time tM = tJ + 2.  From that, we can calculate how fast Mercutio is moving.  We can figure out how far Mercutio travels from his starting point to the y axis, and then we can calculate how long it takes, because we know how fast Mercutio is moving.  And then we're done!

 

Ok, Juliet travels from (0, -6) to (10, 5) in ten seconds.  The distance dJ between these two points is given by

sqrt( (10-0)^2 + (5 - -6)^2)

= sqrt(10^2 + 11^2)

= 14.87

 

So Juliet is moving at 1.487 units/s. 

 

Well, Juliet left (0, -6), and crossed the x axis somewhere along the line at some point (x, 0).  We want to know where that point is.  The slope of Juliet's line of travel is given by rise over run:

 

m = (5 - -6) / (10 - 0) = 11/10 = 1.1

 

But the slope can also be calculated from another two points.  We'll use (0, -6) and (x,0):

 

1.1 = m = (0 - -6) / (x-0) = 6 / x

 

So multiplying both sides by x, we have

 

x * 1.1 = 6

 

x = 5.455

 

So the point on the x-axis that Juliet passes through is given by (5.455, 0).

 

Ok, let's find the time tJ it takes Juliet to get there.  The distance between (0, -6) and (5.455, 0) is

sqrt ( (5.455-0)^2 + (0 - -6)^2)

= sqrt (5.455^2 + 6^2)

= 8.110

 

Now time is given by distance / speed.  At a rate of 1.487 units per second, we see that

tJ = 8.110 / 1.487

    = 5.454

 

Mercutio passes through this same point, 2 seconds after Juliet. 

tM = 5.454 + 2

     = 7.454

 

To calculate Mercutio's speed, we'll need the distance he traveled during that time.

 

d = sqrt( (7-5.455)^2 + (-18 - 0)^2 )

  = sqrt(2.387 + 324)

  = 18.07

 

Mercutio traveled 18.07 units in 7.545 seconds, giving a rate of

18.07/7.545 = 2.395 units / second

 

What we need now is the distance Mercutio travels to the y axis.  We'll use the same trick as before.

 

m = (0- -18) / (5.455 - 7)

    = 18 / -1.545

    = -11.65

 

But if (0, y) is the point where Mercutio crosses the y axis,

 

m = (y- -18) / (0 - 7)

-11.65 = (y+18) / (-7)

 

Multiplying both sides by -7, we have

81.55 = y+18

Subtracting 18 from both sides, we get

 

63.55 = y

 

So Mercutio crosses at (0, 63.55). 

 

And how far is that?  That distance is given by

 

sqrt( (7-0)^2 + (-18 - 63.55)^2 )

= sqrt( 49 + 6650)

= 81.85

 

And now, happy day, we know the time is given by distance over rate.  Mercutio travels 81.85 at a speed of 2.395 units/s, so the time is given by

 

81.85 / 2.395 = 34.17.

 

And that's it.  Our answer is that Mercutio took 34.17 seconds to reach the y axis.

 

And if by some miracle I haven't messed up with my calculator or made any typos, that should be the right answer. If I messed up the numbers somewhere, I do most sincerely apologize.  Anyway, hope it helps.

 

Cheers,

Mike N.