A speeder passes a parked police car at a constant speed of 28.4 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.93 m/s2. How much time passes before the speeder is overtaken by the police car? Answer in units of s

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A speeder passes a parked police car at a constant speed of 28.4 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.93 m/s2. How much time passes before the speeder is overtaken by the police car? Answer in units of s

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Let's set x = 0 where the police car is. What's the equation of motion for the two cars? \[ x_{speeder}(t) = ... what? \] \[ x_{police}(t) = ... what? \]
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

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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?
The position of the speeder is\[ x_s(t) = 28.4 t \]The position of the police car is \[ x_p(t) = \frac{1}{2} 2.93t^2 \] Hence the positions are equal when \[ 28.4 t = \frac{1}{2}2.93t^2 \] Now solve for t as I've written above. You can factor out a t from both sides if t is not zero, which gives you \[ \frac{1}{2}2.93t = 28.4 \] hence \[ t = \frac{2 \times 28.4}{2.93} \] Now simplify that expression.
19.39
Yes, t = 19.39 seconds
I was making this more difficult by using quadratic equation. Thank you for the help and clarification.

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