# What should I look at? What formula do I use? How should I approach this problem? Any help/advice/tips is appreciated.

Feb 13, 2018

The speeder is caught after 896 m.

#### Explanation:

Data you have: $45 \frac{m}{s} , 5.2 \frac{m}{s} ^ 2 , 3 s$

Those 3 seconds give the speeder a lead. How big a lead?

$\text{lead} = 45 \frac{m}{s} \cdot 3 s = 135 m$

It can simplify a problem like this if you choose well on the reference system. You are a passenger looking back at the police car when the 3 seconds has expired. The police car is -135 m from you and falling further back at -45 m/s, but is accelerating toward you at $+ 5.2 \frac{m}{s} ^ 2$.

Use the formula
$s = {s}_{i} + u \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$

Why does this formula have the ${s}_{i}$ term? That is not usually part of it because usually the starting point is at the origin of the reference system.

$0 = - 135 m - 45 \frac{m}{s} \cdot t + \frac{1}{2} \cdot 5.2 \frac{m}{s} ^ 2 \cdot {t}^{2}$

This is going to be a quadratic equation type solution. Simplify. Drop the units, clearly t will be seconds.

$2.6 {t}^{2} - 45 \cdot t - 135 = 0$

$t = \frac{45 \pm \sqrt{{45}^{2} - 4 \cdot 2.6 \cdot \left(- 135\right)}}{2 \cdot 2.6}$

$t = \frac{45 \pm 58.6}{2 \cdot 2.6}$

$t = \frac{103.6}{5.2} , - \frac{13.6}{5.2} = 19.9 s , - 2.6 s$

Clearly the negative time can be discarded. So it took almost 20 s to catch the speeder.

a) Note: Using the speeder as the origin of the reference system was useful above. Now it is better to use the police car's starting point as the reference.

Both speeder and police car travel same distance when the police catches speeder. Easiest way to get the distance, d, is calculate how far the speeder went in those 19.9 s at the constant speed of 45 m/s.

$d = \text{speed" * "time} = 45 \frac{m}{s} \cdot 19.9 s = 896 m$

b) Plot these expressions:

Speeder's distance, ${d}_{s}$, expression

${d}_{s} = 45 \frac{m}{s} \cdot t$

Police distance, ${d}_{p}$, expression

${d}_{p} = \frac{1}{2} \cdot 5.2 \frac{m}{s} ^ 2 \cdot {t}^{2}$

When the plots get to a distance of 896 m, the chase is complete for both.

I hope this helps,
Steve