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A student leaves the university at noon (12:00) bicycling south at a constant rate. At 12:30PM a second student leaves the same point and heads west, bicycling 7mph faster than the first student. At 2:00 they are 30 miles apart. How fast is each one going?

Thanks
Narumi

2006-09-21 19:28:04 · 9 answers · asked by n4rumi 2 in Science & Mathematics Mathematics

9 answers

Speed = Distance / Time.

S1 → Speed one.
S2 → Speed two.
D1 → Distance one.
D2 → Distance two.
t1 → time one.
t2 → time two.
Let x = Speed 1.

Speed = Distance / Time.
S1 = D1/t1.
Let x = Speed 1.
x = D1/t1
x = D1/ 2
D1 = x/ 2
---------------
Speed = Distance / Time.
S2 = D2/t2.
Let (x + 7) = Speed 2.
x + 7 = D2/t2
x + 7 = D2/ [3/2]
D2 = (3/2)(x + 7)
D2 = 3/2x + 21/2
-------------------------

(30)² = (D1)² + (D2)²
(30)² = (2x)² + (3/2x + 21/2)²
(30)² = (2x)(2x) + (3/2x + 21/2)(3/2x + 21/2)
(30)² = 4x² + 9/4x² + 63/4x + 63/4x + 441/4
(30)² = 4x² + 9/4x² + 126/4x + 441/4
(30)² = 16/4x² + 9/4x² + 126/4x + 441/4
(30)² = 25/4x² + 126/4x + 441/4
900 = 6∙25x² + 31∙5x + 110∙25
6∙25x² + 31∙5x + 110∙25 - 900 = 0
6∙25x² + 31∙5x - 789∙75 = 0 (÷ 0∙25)
25x² + 126x - 3159 = 0 Quadratic equation.

[-b±√(b² - 4ac)] / 2a

-(126) ± √[(126)² -4(25)(-3159)] / 2(25)
-(126) ± √[15876 +315900] / 50
-(126) ± √331776 / 50
-(126) ± 576 / 50
- 702/50 (or) 450/50
Negative speed or 9 mph.

S1 = D1/t1.
9 = D1/2
9*2 = D1
D1= 18 miles.

S2 = D2/t2.
x + 7 = D2/t2
9 + 7 = D2/[3/2]
16(3/2) = D2
D2 = 24

(30)² = (D1)² + (D2)²
(30)² = (18)² + (24)²
(30)² = 324 + 576
900 = 900 (Answer confirmed).

S1 = 9 mph.
S2 = x + 7
S2 = 16 mph.

2006-09-21 21:45:03 · answer #1 · answered by Brenmore 5 · 0 0

Here are set-ups for these problems #91 Reciprocal means that if you multiply the two together, you get 1 (x+2)(x-2) = 1 x^2 - 4 = 1 x^2 = 5 #92 (x/8)+(3/x)=7/4 You want to eliminate the x in the denominator. Multiply through by x (x^2 / 8)+ 3 = 7x/4 Since there are no fractions in any of the possible answers, multiply through by 8 x^2 + 24 = 14x #94 First notice that among the possible answers there are some that cannot be right. Substituting 0 for y would give you a division by 0 in the (20/y) term. Also Substituting -9 would give division by 0 in the 20/(y+9) term. If you wanted to try the substitutions, you would only have to try -5 and 36 or you could solve the problem. As with #92, eliminate the fractions. First multiply by y and then by (y+9) 20+ (20y/(y+9)) = y 20(y+9) + 20y = y(y+9) 20y + 180 + 20y = y^2 + 9y 0 = y^2 - 31y - 180 You can use the quadratic formula here or you can factor. #124 Multiply to eliminate the fraction For the moment, ignore the absolute value signs, just treat it as another set of parentheses (2x-1)=3(x-3) 2x-1 = 3x -9 8 = x If you try substituting 2 into the equation, you have (2*2-1)/(2-3) = (4 - 1)/(-1) = 3/(-1) = -3 So B is the only true solution

2016-03-27 01:53:16 · answer #2 · answered by Anonymous · 0 0

Let the student who leaves at 12.00 be student A
at 12.30 student B

the speed of A = x mph
B= ( x+7)mph

cycling at different directions the make up a triangular cycle. having an hypotenuse of 30 miles. so we hav to use pythogras theoreum
A^2 +B^2 = C^2

to use this we have to convert all our terms to b alike. meaning look for distance. Distance = Speed * time

For student A: Time = 1400hrs - 1200hrs = 0200 hrs
Distance= 2x

B= 1400hrs-1200hrs = 1hr 30mins (be careful wen substracting hrs n mins)
which is the same as 3/2
Distance =3/2(x+7)

to solve for x = ( 3/2(x+7) )^2 + (2x)^2 = 30^2

x = 9
so the speed of A = 9mph
B = 9 +7 = 16mph

I hope u understand th working

2006-09-21 21:00:49 · answer #3 · answered by smart_eluh 4 · 0 0

Let
x = speed of bike leaving at noon,
x+7 = speed of bike leaving at 12:30

Then,
(2*x)^2 + ((x+7)*1.5)^2 = 30^2

since d = r*t (distance = rate * time),
bike leaving at noon travels for 2 hours,
bike leaving at 12:30 only travels for 1.5 hours,
use pythagorean theorem, x^2 + y^2 = z^2.

Collecting terms gives the quadratic equation,
25*x^2 + 126*x -3159 = 0.

Plugging coefficients into quadratic equation gives,
x = 9. So speed of other bike is,
x+7 = 16

2006-09-21 20:34:52 · answer #4 · answered by Joe C 3 · 0 0

let s be the speed of the first student
s+7 = speed of second student

(s(2))^2 + ((s+7)(1.5))^2 = 30^2
solve for s

2006-09-21 19:44:13 · answer #5 · answered by Anonymous · 0 0

3X-4Y = 42
X^2 + y^2 = 900

solve the above 2 equations 2 get the answer.

2006-09-21 19:50:03 · answer #6 · answered by hari_mpkumar 1 · 0 1

First one is with speed 15mph
second one is with speed 20mph

2006-09-21 19:42:42 · answer #7 · answered by Promise of the Storm 2 · 0 0

let the speed be x.then
d1=x*2

d2=(x+7)*1.5

d1-d2=30
or, 2x -1.5x-10.5=30
or, 0,5x=40.5
so, x=81 mph
the speeds are 81 mph and 88 mph

2006-09-21 20:15:38 · answer #8 · answered by Anonymous · 0 1

always been lousy at math but I smell a hypotenuse

2006-09-21 19:39:25 · answer #9 · answered by Anonymous · 0 1

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