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the speeder's location xs(t) is
xs(t) = 28.4t
Now what's the police car's position, xp(t) ?

(1/2)(2.93)t^2

Right, exactly. Now when are the two equal?

set them = to each other?

when the police car catches the speeder

Set them equal and find the value of t for which that is the case.

48.46 and 0

obviously its not 0

Right. t = 0 is when the driver first goes past the police car. You want the other answer.

and it is not correct. I've already tried it previously.

t = 48 sec isn't right. Check your calculation

\[ 28.4 t = \frac{2.93}{2} t^2 \]

\[ t = 0 \]
or
\[ t = \frac{2 \times 28.4}{2.93} \]

I made an error somewhere, thank-you.

2.93? wait

2 * a
a=.586?

.586^2 ?

noooo

1/2 * 2.93?

19.39

Yes, t = 19.39 seconds