<p>check this problem :
A car traveling with a constant speed of 30m/s pass a police officer located behind a billboard. One second (after this happens) the police chases the car with a constant acceleration of 3m/s^2. How long does it Take the police to undertake the speeding car?
with Detailed explanation and thank</p>
<p>x is position of speeding car relative to billboard.
y is position of police car relative to billboard.</p>
<p>t is the time measured from when the speeding car passes the dashboard.</p>
<p>x=vt=30t
y= (1/2) 3<em>(t-1)</em>(t-1) [note that the initial velocity of the police car is 0 and the initial position of the police car is 0 and the acceleration is 3). The police car starts moving at t=1, hence the (t-1) in the equation. I assume you know the equation for distance travelled under constant acceleration — initial_position + initial-veleocity<em>time + (1/2) acceleration</em>time^2</p>
<p>The police car reaches the speeding car when x=y</p>
<p>30t = (1.5)<em>(t-1)</em>(t-1)</p>
<p>Solve the quadratic for t. Then subtract 1 since the police car starts at t=1. I don’t have a calculator with me. The answer is about 19 seconds.</p>
<p>thanks 4 the help</p>