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?

Since the skiers are travelling in perpendicular directions, the final distance between them can be caculated using Pythagorean's Theorem: a^{2} + b^{2} = c^{2}

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

Answered By Marija J

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) $d^2=x^2+y^2$d^{2}=x^{2}+y^{2}

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=v_{x}(t+16) , and

(3) $y=v_yt$y=v_{y}t ,

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

and v_{x} = 36 km/h and v_{y} = 44 km/h. Since the required final distance is d = 20 km, we can rewrite equation (1) as

The above involves squaring the binomial (t+1/6)^{2 }= t^{2} + 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.$808t^{2}+108t−91=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 is17 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=√15.9^{2}+12.1^{2}=√399.22=19.98≈20 km.

3 years ago

## Answered By Deeti P

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.

$v_1=\frac{\bigtriangleup x_1}{\bigtriangleup t_1}$v

_{1}=?x_{1}?t_{1}$v_2=\frac{\bigtriangleup x_2}{\bigtriangleup t_2}$v

_{2}=?x_{2}?t_{2}Since the skiers are travelling in perpendicular directions, the final distance between them can be caculated using Pythagorean's Theorem: a

^{2}+ b^{2}= c^{2}In this case,

$\bigtriangleup x_1^2+\bigtriangleup x_2^2=\bigtriangleup x^2$?x

_{1}^{2}+?x_{2}^{2}=?x^{2}Rearranging the motion equations to isolate for displacement, we have:

$\bigtriangleup x_1=v_1\bigtriangleup t_1$?x

_{1}=v_{1}?t_{1}$\bigtriangleup x_2=v_2\bigtriangleup t_2$?x

_{2}=v_{2}?t_{2}Let's put these equations into Pythagorean's Theorem.

$\left(v_1\bigtriangleup t_1\right)^2+\left(v_2\bigtriangleup t_2\right)^2=\bigtriangleup x^2$(v

_{1}?t_{1})^{2}+(v_{2}?t_{2})^{2}=?x^{2}Now let's figure out the changes in time:

$\bigtriangleup t_1=\left(t_f-t_o\right)_1=t_f-0=t_f$?t

_{1}=(t_{ƒ }−t_{o})_{1}=t_{ƒ }−0=t_{ƒ }$\bigtriangleup t_2=\left(t_f-t_o\right)_2=t_f-10min\left(\frac{1hr}{60min}\right)$?t

_{2}=(t_{ƒ }−t_{o})_{2}=t_{ƒ }−10min(1hr60min )So then,

$\left(v_1t_f\right)^2+\left(v_2\left(t_f-\frac{1}{60}\right)\right)^2=\bigtriangleup x^2$(v

_{1}t_{ƒ })^{2}+(v_{2}(t_{ƒ }−160 ))^{2}=?x^{2}Now all values are known except for t

_{f}. Rearrange and solve.*Note that your answer will be 8 a.m. plus t

_{f}3 years ago

## Answered By Deeti P

*My last equation should have read:

$\left(v_1t_f\right)^2+\left(v_2\left(t_f-\frac{1}{6}\right)\right)^2=x^2$(v

_{1}t_{ƒ })^{2}+(v_{2}(t_{ƒ }−16 ))^{2}=x^{2}(I apologize for all of the question marks in the equations...they are meant to be deltas, ie. change in)

3 years ago

## Answered By Negar S

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 :) )

a

^{2}+b^{2}=20^{2}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}+b^{2}=20^{2}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:

Play Drawing3 years ago

## Answered By Negar S

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

## Answered By Marija J

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) $d^2=x^2+y^2$d

^{2}=x^{2}+y^{2}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=v

_{x}(t+16 ) , and(3) $y=v_yt$y=v

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

and v

_{x}= 36 km/h and v_{y}= 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$20

^{2}=(36·(t+16 ))^{2}+(44t)^{2}.The above involves squaring the binomial (t+1/6)

^{2 }= t^{2}+ 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.$808t

^{2}+108t−91=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=√15.9

^{2}+12.1^{2}=√399.22=19.98≈20 km.