Two cars leave an intersection at the same time. One drives east while the other travels south at 15 miles per?

October 28th, 2009 | by Michael |
cars
ERIK asked:


Two cars leave an intersection at the same time. One drives east while the other travels south at 15 miles per hour faster than the other. After 3 hours, the cars are 225 miles apart. How fast is the southbound car driving?

SIDNEY

  1. 2 Responses to “Two cars leave an intersection at the same time. One drives east while the other travels south at 15 miles per?”

  2. By cidyah on Nov 1, 2009 | Reply

    Let the speed of car driving east be x mph.
    Then, the speed of car driving south is x+15 mph
    In 3 hours, the car driving east has traveled 3x miles
    In 3 hours, the car driving south has traveled 3(x+15) = 3x+45 miles
    (3x)^2 +(3x+45)^2 = 225^2 (Pythogorean theorem)
    9x^2 + 9x^2+270x+2,025 = 50,625
    18x^2 +270x-48,600=0
    This equation is of form ax^2+bx+c
    a = 18 b = 270 c = -48600
    x=[-b+/-sqrt(b^2-4ac)]/2a]
    x=[-270 +/-sqrt(270^2-4(18)(-48600)]/(2)(18)
    discriminant is b^2-4ac =3572100
    x=[-270 +?(3572100)] / (2)(18)
    x=[-270 -?(3572100)] / (2)(18)
    x=[-270+1890] / 36
    x=[-270-1890] / 36
    The roots are 45 and -60
    x=45 mph driving east
    =60 mph drving south

  3. By illusionk on Nov 3, 2009 | Reply

    EDIT: Ok I submitted late, and the guy who submitted before me is obviously right as well. His method was similar yet different. Some of the actual equations he used were obtained using Calculus concepts. Mine uses pythagorean theorem and speed formula with some obvious logic.

    The answer is 60mph south. For the eastbound car obviously 45 miles an hour.

    To solve this we will need our equations. We are working with speed, so obviously we need the formula for that.

    Speed = Change in Distance (D) / change in time

    Ok now make a diagram. We draw the two cars heading in their respective directions and we see it forms a right triangle. We see the distance between them is the hypotneuse which is given by the pythagorean theorem which I really hope you know.
    a^2 + b^2 = c^2
    We will have a be the distance traveled for the eastbound car, and b for the southbound car.
    The distance between them c is 225.

    so,
    a^2+b^2 = 225^2 = 50625

    Now we have the distance between them related to two currently unknown distances traveled. However we can represent each distance traveled with an equation related to the speed the car a or b is traveling.

    Speed = distance/time
    so distance traveled = Speed*time

    We aren’t given the exact speeds, because if we were the problem would have taken like 2 secs to solve. But we are given the speeds in relation to each other.
    We don’t know what SpeedA is…it’s just some number
    SpeedB = 15 mph faster than SpeedA

    Now back to our pythagorean equation. We have a^2 and b^2…these are distances not speeds. So we represent the distances using the formula we derived above, distance = speed*time
    SpeedA is just whatever SpeedA is…we want to find this number out, since we don’t know it yet.
    SpeedB = SpeedA + 15

    so
    DistanceA = SpeedA*3 [3hrs]
    DistanceB = SpeedB*3= (SpeedA+15)*3

    replace those in our equation.
    a^2+b^2=50625
    (SpeedA*3)^2 + ((SpeedA+15)*3)^2 = 50625
    Use simple algebra and solve
    SpeedA = 45
    Now just add 15 to it to get the other speed
    SpeedB=60

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