Find the parametric equation of the path of an object going at a speed x , with an orientation vector at time t and point p

$$x = 1/2, v[2,−2], t = 3, P(1, 2)$$ x = speed, v = orientation, t = time, p is a point.

I tried this : $$((2, -2) - (1,2)) \sqrt{1^2 + 4^2}$$

I got my <1,4> from $(2-1, 2-(-2))$ , which is my directional vector. Right ?

which is equal to $$(1, -4) \sqrt{17}$$

my final answer is : $$(1,2) + 1/2((1,-4)\sqrt {17})$$ Is it ok ?

Then I'm asked : what would be the equation if it was 'unit speed' instead of m/sec

Not sure what to do.

Thanks

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Assuming linear motion with no acceleration:

First, let's normalize(make length 1) the direction vector v:

$$v = <2, -2>$$

Dividing by the length, $\sqrt{(-2)^2 + (2)^2}$, we get:

$$\hat v = 1/\sqrt{8}<-2,2> = 1/\sqrt{2}<-1,1>$$

Now we can plug this in to the linear equation $(x,y) = \hat v t + (x_0,y_o)$

$$(1,2) = (1/\sqrt{2}<-1,1>)3 + (x_0,y_o)$$

Rearranging, we get:

$$(x_0,y_o) = (1+3/\sqrt{2}, 2-3/\sqrt{2})$$

Plugging this back into out equation:

$$(x,y) = (1/\sqrt{2}<-1,1>) t + (1+3/\sqrt{2}, 2-3/\sqrt{2})$$

In parametric form:

$$x(t) = -t/\sqrt{2} + (1 + 3/\sqrt{2})$$ $$y(t) = t/\sqrt{2} + (2 - 3/\sqrt{2})$$

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