1) The police officer catches the speeder after a time of 19.93 s (measured from the moment the speeder passes the officer)
2) The officer catches the speeder after a distance of 299 m
Explanation:
1)
The speeder is driving at constant speed, so we represent the position of the speeder at time t as
where
is the speed of the speeder
t is the time measured from the moment the speeder passes the officer
The police car starts its motion after 9 seconds, driving from rest and accelerating at , so its position at time t can be written as
where
is the acceleration of the car
The police officer catches the speeder when the two positions are equal, so:
And solving for t,
which gives two solutions:
t = 4.06 s
t = 19.93 s
However, the time must be larger than 9 seconds (because we are measuring the time from the instant the speeder passes the police officer, and the officer starts its motion only 9 seconds later), so the correct solution is
t = 19.93 s
2)
To find how far the police officer went before the speeder was caught, we just substitutite t = 19.93 s into its equation of motion.
We find:
And we can verify that the speeder covered the same distance in this same time:
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