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Diano4ka-milaya [45]
3 years ago
5

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

iform acceleration of 2.75 m/s 2 . How much time passes before the speeder is overtaken by the police car? Answer in units of s.
Physics
1 answer:
KonstantinChe [14]3 years ago
7 0

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:

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

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