- #1

i_love_science

- 80

- 2

- 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.

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.