# Cops vs speeder - time

Tags:
1. Sep 8, 2015

### Rectifier

This problem was translated from Swedish, sorry for any grammatical errors present.

1. The problem
A cop sets up a radar trap. A speeder passes the policeman and continues going at $60km/h$ even though the maximal allowed speed is $50km/h$. The cop starts accelerating at $5 m/s^2$ directly (without reaction-time) . The speeder does not notie anything and continues at the same speed.

a) how many seconds does it take for the policeman to reach the speeder
b) If the speeder has a constant velocity v and the cop has a constant acceleration a, find an expression for the time it takes for the cop to reach the speeder.
c) Now, assume that the cop ( in a) ) has a reaction time of $1$ second (chase starts one second later). Find a quadratic equation and solve it to find the time it takes for the cop to reach the speeder.​

2. Relevant equations
$d = v_0t + \frac{at^2}{2}$
1km = 1000m
1h = 3600secs

3. The attempt
a)

With the help of $d = v_0t + \frac{at^2}{2}$ I construct two equations:
$d_1 = v_0t + \frac{at^2}{2}=60km/h \cdot t \ s = \frac{60}{3.6}m/s \cdot t \ s \\ d_2 = \frac{5t^2}{2}$

the cop reaches the speeder when $d_1 = d_2$
$d_1 = d_2 \\ \frac{60}{3.6}t = \frac{5t^2}{2} \\ \frac{5t^2}{2}-\frac{60}{3.6}t = 0 \\ t(\frac{5t}{2}-\frac{60}{3.6}) = 0 \\ t_1 = 0 \\ t_2: \\ \frac{5t}{2}- \frac{60}{3.6}=0 \\ t_2=\frac{2 \cdot 60}{3.6 \cdot 5}=6.667... \ sec$

The cop reaches speeder at t=6.667. $t_1$ is the moment when speeder passes the cop in the beginning.

b) expression
$\frac{vt}{3.6} = \frac{at^2}{2} \\ \frac{at^2}{2}-\frac{vt}{3.6} = 0 \\ t(\frac{at}{2}-\frac{v}{3.6}) = 0 \\ t_1 = 0 \\ t_2 : \\ \frac{at}{2} = \frac{v}{3.6} \\ t_2 = \frac{2v}{3.6a} =\frac{v}{1.8a}$

A test for v=60 and a=5 gives us:
$t_2 = \frac{v}{1.8a} = \frac{60}{1.8 \cdot 5}=6.667...$

c)
The policeman starts 1 second later. The graph would then move one second to the right.

The equation for the cop is therefore

$d_1 = v_0t + \frac{at^2}{2}=60km/h \cdot t \ s = \frac{60}{3.6}t \ s \\ d_2 = \frac{5(t-1)^2}{2}$

The equation for the speeder stays the same

$d_1 =d_2$ gives

$d_1 = d_2 \\ \frac{60}{3.6}t = \frac{5(t-1)^2}{2} \\ \frac{5(t-1)^2}{2} - \frac{60}{3.6}t = 0 \\ (5 t^2)/2-5 t+5/2- \frac{60}{3.6}t = 0 \\$
Wolfram gives me $t_1 = 0.116963$ and $t_2 = 8.5497$

I was expecting to get $t_1 = 0$ and t_2 like $7.7$ or something.

2. Sep 8, 2015

### CWatters

Have a think about that bit again. You have added subtracted 1 second to the part of the equation that deals with the acceleration of the cop. That's not correct, he doesn't accelerate for an extra second less time. The extra second occurs before he starts to accelerate.

Last edited: Sep 8, 2015
3. Sep 8, 2015

### Rectifier

So in other words I should not count with t_1 since its <1, am I right?

4. Sep 8, 2015

### CWatters

Not sure I understand your #3

I would have written...

d1 = distance travelled in one second + distance travelled in the time t