We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

$x=vt=23.3\frac{m}{s}t$

and the police car distance:

$x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}$

Since they both travel the same distance x, we can equal both formulas and solve for t:

$0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\$

Two solutions exist to the equation; the first one being $t=0$

The second solution will be:

$0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s$

This result allows us to confirm that the police car will take 32s to catch up to the speeder

7 0

2 years ago

A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires

Olegator [25]

Now I can actually edit my answer directly: I'm fairly sure I've got this wrong, and my mind has gone blank for how to do it, if someone could delete this that would be great and I'll think about it and see if I can figure it out!

4 0

3 years ago