Speed of car = 130 km/hr = 130*5/18 = 36.11 m/s
Let the time taken for catching up be t,
Distance traveled by car = 36.11 t
Acceleration of police vehicle = 10.1 km/h/s = 10.1*5/18 = 2.81 m/[tex]s^2[/tex]
Distance traveled by police vehicle in time t = [tex]ut+\frac{1}{2} at^2[/tex]
Here u = 0, a = 2.81 m/[tex]s^2[/tex]
So, Distance = 0*t+ [tex]\frac{1}{2} 2.81*t^2=1.405t^2[/tex]
Comparing both the distances we have
[tex]36.11 t = 1.405t^2\\ \\ t = 25.70 seconds[/tex]
So after 25.70 seconds police vehicle will catch car.
The time is taken by the police to catch the car as the driver passes the officer is 25.793 sec.
Given to us
Speed of the car = 130 km\h = 36.11 m/sec
Acceleration of the police car = 10.1 km/h = 2.80 m/sec²
Assumption
Let the time taken by the police to catch the car be t.
The distance traveled by the police is the same as the distance traveled by car after police start chasing them. therefore,
Distance = velocity of the car x time
S = 36.11 t
[tex]S=ut +\dfrac{1}{2}at^2[/tex]
Substitute the values,
[tex]36.11t=0 +\dfrac{1}{2}(2.80)t^2[/tex] {the car was stationary at the beginning, u=0},
[tex]t = 25.793\ sec[/tex]
Hence, the time taken by the police to catch the car is 25.793 sec.
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