An unmarked police car travelling at a constant 80.0 km/h is passed by a speeder travelling at 100.0 km/h. Precisely 1.0 s after the speeder passes, the police steps on the accelerator. If the police car’s acceleration is 2.0 m/s2, how much time passes before the police car overtakes the speeder (assumed moving at a constant speed)?

Mar 8, 2017

$6.4 s$, rounded to one decimal place.

Explanation:

Let the Police car overtake the speeder after a period $t$ passes once he crosses the police car.

1. For first $1.0 s$ both cars travel at their respective speeds. Distance between the two cars at the end of $1.0 s$
${d}_{1} = \text{Relative speed of cars" xx"1.0}$
${d}_{1} = \left[\left(100.0 - 80.0\right) \times \frac{1000}{3600}\right] \times \text{1.0} = \frac{20000.0}{3600} = \frac{50}{9} m$
This is the distance speeder is ahead of police car just before it accelerates.
2. Distance moved by speeder after $1.0 s$ before Police car overtakes it
${d}_{2} = 100.0 \times \frac{1000}{3600} \times \left(t - 1\right) = \frac{250}{9} \left(t - 1\right) m$
3. Distance to be covered by Police car to overtake the speeder in $\left(t - 1\right) s$ is
${d}_{1} + {d}_{2} = \frac{50}{9} + \frac{250}{9} \left(t - 1\right) m$
4. Setting up the kinematic expression
$s = u t + \frac{1}{2} a {t}^{2}$
and inserting given values in SI units we get
$\frac{50}{9} + \frac{250}{9} \left(t - 1\right) = \frac{200}{9} \left(t - 1\right) + \frac{1}{2} \times 2.0 {\left(t - 1\right)}^{2}$
$\implies {\left(t - 1\right)}^{2} - \frac{50}{9} \left(t - 1\right) - \frac{50}{9} = 0$
$\implies 9 {\left(t - 1\right)}^{2} - 50 \left(t - 1\right) - 50 = 0$
To solve the quadratic we substitute $t - 1 = x$
It becomes
$9 {x}^{2} - 50 x - 50 = 0$

I found roots of this quadratic using inbuilt graphic tool we get roots as $x = - 0.9 \mathmr{and} 6.4$, rounded to one decimal place
Time can not be negative, therefore, ignoring the first root we get

$\implies t - 1 = 6.4$
$\implies t = 6.4 s$, rounded to one decimal place.