Answer:
32s
Explanation:
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:
[tex]x=vt=23.3\frac{m}{s}t[/tex]
and the police car distance:
[tex]x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}[/tex]
Since they both travel the same distance x, we can equal both formulas and solve for t:
[tex]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} )\\\\[/tex]
Two solutions exist to the equation; the first one being [tex]t=0[/tex]
The second solution will be:
[tex]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[/tex]
This result allows us to confirm that the police car will take 32s to catch up to the speeder
