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? 


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Erdos Erdos
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