$How do you go about solving this problem?$

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$How do you go about solving this problem?$

A particle P is initially at the point with position vector (30, 10) and moves with a constant speed of 10ms^-1 to the same direction as (-4, 3) to find the position vector of p after t seconds.

I've tried solving it like this but I get the completely wrong answer

$\sqrt{(-4-30)^2+(3-10)^2} =\sqrt{1205} $

$\binom{-34\div \frac{\sqrt{1205} }{10^2}}{-7\div \frac{\sqrt{1205} }{10^2}} $

My idea was that If you find the ratio difference between the length of (30, 10) to (-4, 3) = $\sqrt{1205} $ and $10ms^{-1}$ you could then use that ratio to find the change in the x and y value.

I'm not what I am doing wrong but I am assuming that I am understanding the question incorrectly?

Vectors

Math Gnome

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You fist need to find a unit vector in the direction of $(-4,3)$:
\[u=\frac{(-4,3)}{\sqrt{(-4)^2+3^2}}=\frac{(-4,3)}{\sqrt{16+9}}=\frac{(-4,3)}{\sqrt{25}}=\frac{(-4,3)}{5}.\]

After $t$ seconds, the displacement of the particle is
\[D(t)=t\times u=t \times \frac{(-4,3)}{5}=\frac{(-4,3)t}{5}.\]
The position of the partcile is given by
\[\text{Position}=\text{Initial Point}+\text{Displacement}\]
Hence the position after $t$ seconds is given by
The position of the particile after time $t$ is given by the formula
\[P(t)=(30,10)+\frac{(-4,3)t}{5}=(30-\frac{4t}{5}, 10+\frac{3t}{5}).\]

Erdos

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