# Vectors - finding coordinates of collision point

• i_love_science
In summary: But that's a bit of a "trick" that isn't always obvious, and it doesn't generalize well.Instead, the solution is showing you a more general technique. You start with the parametric equations$$x = x_0 + at \\ y = y_0 + bt$$where ##a## and ##b## are constants. You want to get an equation for ##y## in terms of ##x##. So you solve the first equation for ##t##:$$t = \frac{x - x_0}{a}$$Substitute this into the second equation:$$y = y_0 + b \frac{x - x_0}{a}$$This simpl
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.

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##.

i_love_science and SammyS

## 1. How do vectors help in finding the coordinates of a collision point?

Vectors are mathematical quantities that have both magnitude and direction. When two objects collide, they exert forces on each other in different directions. By using vector addition and subtraction, we can determine the overall force acting on each object and calculate the coordinates of the collision point.

## 2. What are the steps involved in finding the coordinates of a collision point using vectors?

The first step is to draw a diagram of the collision, with the initial positions and velocities of the objects involved. Then, we use vector addition and subtraction to calculate the overall force acting on each object. Next, we use this force to determine the acceleration of each object. Finally, we can use the equations of motion to find the coordinates of the collision point.

## 3. Can vectors be used to find the coordinates of a collision point in 3-dimensional space?

Yes, vectors can be used to find the coordinates of a collision point in any number of dimensions. The principles of vector addition and subtraction remain the same, but the calculations may be more complex in higher dimensions.

## 4. Are there any limitations to using vectors to find the coordinates of a collision point?

Vectors can only be used to find the coordinates of a collision point if the collision is a result of two objects exerting forces on each other. If the collision is caused by other factors, such as friction or external forces, vectors may not be applicable.

## 5. How can the accuracy of using vectors to find the coordinates of a collision point be improved?

The accuracy of using vectors to find the coordinates of a collision point can be improved by using more precise measurements of the initial positions and velocities of the objects involved. Additionally, using more advanced mathematical techniques, such as calculus, can also improve the accuracy of the calculations.

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