Vectors - finding coordinates of collision point

AI Thread Summary
The discussion focuses on finding the collision point of two cars using their parametric equations. The initial calculations yielded incorrect time values for the cars, leading to an incorrect collision point. The correct approach involves finding the normal vector of car 1's direction to derive its Cartesian equation, which is x - 0.8y + 1 = 0. A correction to the parametric equation for car 2 also led to the correct collision point of (5,5). The method of using the normal vector is a standard technique for converting parametric equations into Cartesian form.
i_love_science
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Homework Statement
At noon, Car 1 moves in a straight line, starting at point A (1, 0). Its position t seconds after it starts is given by (x,y) = (1,0) + t (0.8, 1). Car 2 moves with constant speed starting at point B (0,2) and travels in a straight line, given the equation y = 0.6x + 2, x ≥ 0 . Eventually, the two vehicles collide. Find the coordinates of the collision point.
Relevant Equations
vector equation
cartesian equation
For car 1, the parametric equations are x = 1 + 0.8t and y=t. For car 2, the parametric equations are x=0.6s and y=2+s. (Let t and s represent time). Solving the system of equations, when the x values are equated are the y values are equated, I get s = -13 and t = -11. I assume that the 2 cars don't start moving at the same time, since the time values are different. Substituting t = -11 into the parametric equations for car 1 gives (x,y) = (-7.8, -11) (substituting s = -13 into the parametric equations for car 2 would give me the same result).

My answer was wrong though. The solution says that find the normal vector of (0.8, 1), which is (1, -0.8). Then the scalar equation of the line is x - 0.8y + c = 0, substituting point (1,0) into that gives c= -1. Thus the cartesian equation for car 1 is x - 0.8y + 1 = 0. Solving the system of cartesian equations of car 1 and car 2 gives (x,y) = (5,5).

I don't understand where I went wrong, and why the solution says to find the normal vector to find the cartesian equation for car 1. Thanks.
 
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Your parametric equation for ##y = 0.6x + 2## is not right.
 
PeroK said:
Your parametric equation for ##y = 0.6x + 2## is not right.
Thanks, I fixed it to be x=s and y=2 + 0.6s, and I got the right answer.

But could you explain why the solution says to find the normal vector to find the cartesian equation for car 1?
 
i_love_science said:
Thanks, I fixed it to be x=s and y=2 + 0.6s, and I got the right answer.

But could you explain why the solution says to find the normal vector to find the cartesian equation for car 1?
That's a technique for changing a parametric equation into a Cartesian form. Note that, instead, you could simply substitute ##y = t## into ##x = 1 + 0.8t##.
 
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