Vectors - finding coordinates of collision point

[i_love_science]

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.

 

[PeroK]

Your parametric equation for $y = 0.6x + 2$ is not right.

 

[i_love_science]

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?

 

[PeroK]

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 $x = 1 + 0.8t$.