## Alberta Free Tutoring And Homework Help For Math 20-2

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### A jet skier leaves a dock at 8 a.m. and travels due west at 36 km/h. A second jet skier leaves the same dock 10 min later and travels due south at 44 km/h. At what time of day, to the nearest minute, will the two jet skis be 20 km apart?

3 years ago

Let's use the equation for uniform motion:

$v=\frac{\bigtriangleup x}{\bigtriangleup t}$v=?x?t

Using subscripts to differentiate between the first jet skier (1) and the second jet skier (2), we have the following equations.

$v_1=\frac{\bigtriangleup x_1}{\bigtriangleup t_1}$v1=?x1?t1

$v_2=\frac{\bigtriangleup x_2}{\bigtriangleup t_2}$v2=?x2?t2

Since the skiers are travelling in perpendicular directions, the final distance between them can be caculated using Pythagorean's Theorem: a2 + b2 = c2

In this case,

$\bigtriangleup x_1^2+\bigtriangleup x_2^2=\bigtriangleup x^2$?x12+?x22=?x2

Rearranging the motion equations to isolate for displacement, we have:

$\bigtriangleup x_1=v_1\bigtriangleup t_1$?x1=v1?t1

$\bigtriangleup x_2=v_2\bigtriangleup t_2$?x2=v2?t2

Let's put these equations into Pythagorean's Theorem.

$\left(v_1\bigtriangleup t_1\right)^2+\left(v_2\bigtriangleup t_2\right)^2=\bigtriangleup x^2$(v1?t1)2+(v2?t2)2=?x2

Now let's figure out the changes in time:

$\bigtriangleup t_1=\left(t_f-t_o\right)_1=t_f-0=t_f$?t1=(tƒ to)1=tƒ 0=tƒ

$\bigtriangleup t_2=\left(t_f-t_o\right)_2=t_f-10min\left(\frac{1hr}{60min}\right)$?t2=(tƒ to)2=tƒ 10min(1hr60min )

So then,

$\left(v_1t_f\right)^2+\left(v_2\left(t_f-\frac{1}{60}\right)\right)^2=\bigtriangleup x^2$(v1tƒ )2+(v2(tƒ 160 ))2=?x2

Now all values are known except for tf. Rearrange and solve.

3 years ago

*My last equation should have read:

$\left(v_1t_f\right)^2+\left(v_2\left(t_f-\frac{1}{6}\right)\right)^2=x^2$(v1tƒ )2+(v2(tƒ 16 ))2=x2

(I apologize for all of the question marks in the equations...they are meant to be deltas, ie. change in)

3 years ago

Let's make it fun :

Jet Skier goes to west we call Westy and Jet Skier going to South let's call it Southy.

Westy starts at 8 am with V=36km/h

at 8:10 am, Westy is at the distance of :

V=x/t    It means x=V.t

here t=10mins and V is 36km/h .... it means x = 36km/60mins . 10mins= 6km

so we found at 8:10 am Westy is 6km away from the deck.

Southy starts at 8:10 am towards south

Look at the ugly drawing of mine below ( I found it hard to work in teh drawing :) )

a2+b2=202

while we know a is southy and b is westy.

a velocity is 44 and b is 36

can we say a=44/36 .b ?

yes we can

let's replace now :

(44/36 . b )2+b2=202

b=12.66km

and a=44/36 . 12.66 = 15.47 km

again V=x/t

For Southy:

V=44km/hr

x=15.47km

it means t=15.47 / (44km/60min) = 21 mins

That means 8:21 am they are 20 km away.

Is that not so easy ? :)

#### Attached Whiteboard:

3 years ago

now you can easy make it into equations if you want to show up your math profiency... Let me know if you need more help

3 years ago

This question involves motion in two dimensions. The two jet skiers travel in perpendicular directions -- due west and due south. According to Pythagoras' theorem, the distance between them at any given moment is

(1)                           $d^2=x^2+y^2$d2=x2+y2

where x is the distance covered by the first jet skier and y the distance covered by the second jet skier. As both skiers are moving with uniform speeds,  distance covered by each is calculated as a product of  that skier's speed and total time of motion. The only complication here is that the second skier starts moving ten minutes (or, 1/6 of an hour) after the first one. In other words, the second skier moves for a total time t (in hours), while the first skier moves 1/6 hour longer, or (t + 1/6).

Therefore,

(2)                         $x=v_x\left(t+\frac{1}{6}\right)$x=vx(t+16 ) ,          and

(3)                         $y=v_yt$y=vyt ,

where  t is the time elapsed from 8:10 a.m. -- that is, the time after the second skier started --

and vx = 36 km/h and vy = 44 km/h. Since the required final distance is d = 20 km, we can rewrite equation (1) as

(4)                                $20^2=\left(36\cdot\left(t+\frac{1}{6}\right)\right)^2+\left(44t\right)^2$202=(36·(t+16 ))2+(44t)2 .

The above involves squaring the binomial (t+1/6)2 = t2 + 1/3 t + 1/36, and (after expanding all terms and dividing out by a factor 4 -- ask me if you need extra help with this) eventually becomes a quadratic equation for t (in hours):

(5)                                   $808t^2+108t-91=0.$808t2+108t91=0.

Two solutions result: t = 0.275 h and t = - 0.409 h, but only the positive solution is acceptable here. (Negative t would refer to the time in the past,  when the skiers were not moving.)

Since t = 0.275 h = 16.5 min,  or  ~17 min to the nearest minute, the moment when two jet skiers are 20 km apart is  17 minutes after 8:10 a.m., or at 8:27 a.m.

Finally, a quick check: at that time skier one has moved 6 km + 0.275*36 km = 15.9 km due west, while the skier two has gone 0.275*44 km = 12.1 km south. Distance between them is, therefore

(6)    $d=\sqrt{15.9^2+12.1^2}=\sqrt{399.22}=19.98\approx20$d=15.92+12.12=399.22=19.9820 km.