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 #2
avatar+23254 
0

AX = 4

BY = 18

Call  XC = x

then YC = 2x

 

Triangle(AXC)  is a right triangle   --->   AC2  =  AX2 + XC2  --->   AC2  =  42 + x2   --->   AC  =  sqrt(16 + x2)

Triangle(BYC)  is a right triangle   --->   BC2  =  BY2 + YC2  --->   BC2  =  182 + (2x)2   --->   BC  =  sqrt(324 + 4x2)

 

Area( trapezoid(AXYB) )  =  ½·(x + 2x)(4 + 18)  =  33x

Area( triangle(AXC) )  =  ½·4·x  =  2x

Area( triangle(BYC) ) =  ½·18·2x  =  18x

Area( triangle(ABC) )  =  Area( trapezoid(AXYB) )  -  Area( triangle(AXC) )  -  Area( triangle(BYC) )

                                     =  33x - 2x - 18x  =  13x

 

Also:  area( triangle(ABC) )  ½·AC·BC  =  ½· sqrt(16 + x2)·sqrt(324 + 4x2)

                                                               =  ½· sqrt( (16 + x2) · (324 + 4x2) )

                                                               =  ½· sqrt( 4x4 + 388x2 + 5184 )

 

Therefore:     ½· sqrt( 4x4 + 388x2 + 5184 )  =  13x

                          sqrt( 4x4 + 388x2 + 5184 )  =  26x

                                    4x4 + 388x2 + 5184  =  676x

                                     4x4 - 288x2 + 5184  =  0

                                         x4 - 72x2 + 1296  =  0

                                          (x2 - 36)(x2 - 36  =  0

                            (x + 6)(x - 6)(x + 6)(x - 6)  =  0

 

--->   x  =  6

--->   XC  =  x   --->   XC  =  6

--->   YC  =  2x   --->   YC  =  12

--->   AC  =  sqrt(16 + x2)   --->   AC  =  sqrt(52)

--->   BC  =  sqrt(324 + 4x2)  --->   BC  =  sqrt(468)

 

--->  AB2  =  AC2 + BC2   --->   AB2  =  52 + 468  =  520   --->   AB  =  sqrt(520)

May 30, 2020
 #1
avatar+280 
+1

The interval [0,50] would mean that we use numbers (t) between 0 and 50 to replace with x and y. This would be rather difficult to solve since it would involve a lot of work. 

So, to make this easier, I used the numbers given in the multiple choices, 3.2, 4.8, and 2.4. Because t is measured in tenths of seconds, you'd multiply the numbers by ten, making the numbers 32, 48, and 24. (This will make more sense at the end.)

Using these numbers, make a chart representing Ari's and Matthew's path.

 

 

Ari's Path

t x(t) = 36 + (1/6) t y(t) = 24 + (1/8) t
24 36 + (1/6) t =   36 + (1/6) (24) =   40 24 + (1/8) t =   24 + (1/8) (24) =   27
32 36 + (1/6) t =   36 + (1/6) (32) =   41.3333... 24 + (1/8) t =   24 + (1/8) (32) =   28
48 36 + (1/6) t =   36 + (1/6) (48) =   44 24 + (1/8) t =   24 + (1/8) (48) =   30

 

Matthew's Path

t x(t) = 32 + (1/4) t y(t) = 18 + (1/4) t
24 32 + (1/4) t =   32 + (1/4) (24) =   38 18 + (1/4) t =   18 + (1/4) (24) =   24
32 32 + (1/4) t =   32 + (1/4) (32) =   40            18 + (1/4) t =   18 + (1/4) (32) =   26
48 32 + (1/4) t =   32 + (1/4) (48) =   44 18 + (1/4) t =   18 + (1/4) (48) =   30

 

 

 

Next, you would identify if Ari and Matthew collide by looking at the end results of the x and y. If NONE of the results match, then do NOT collide. If they do collide, you can determine the answer by simply dividing (t) by ten to get your final answer. (t) is the number in red.

 

The end results show that Ari and Mattew do collide at (44, 30). (t) is 48. 

Because t is measured in tenths of seconds, you would divide 48 by 10.

48 / 10 = 4.8

Your answer for when they would collide would then be 4.8. 

 

Ari and Matthew collide at 4.8 seconds.

May 30, 2020
 #1
avatar+284 
+2
May 30, 2020
 #2
avatar+159 
+1
May 30, 2020
 #1
avatar+33661 
+2
May 30, 2020
 #1
avatar+33661 
+4
May 30, 2020
 #2
avatar+223 
-1
May 30, 2020

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