(a) The time it takes for the police officer to catch up to the speeding car is determined as 0.31 s.
(b) The speed of the police officer at the time he catches up to the driver is 136.8 km/h.
<h3>
Time of motion of the police</h3>
The time taken for the police to catch up with the driver is calculated as follows;
v = at
where;
- a is acceleration = 11.8 km/h/s, = 3.278 m/s²
- v is velocity = 135 km/h = 37.5 m/s
t = v/a
t = 37.5/3.278
t = 11.4 seconds
(v1 - v2)t = ¹/₂at² --- (1)
(v1 - v2)t = v1²/2a --- (2)
From (1):
(v1 - 37.5)t = ¹/₂(3.278)t²
(v1 - 37.5)t = 1.639t²
v1 - 37.5 = 1.639t
v1 = 1.639t + 37.5 -----(3)
From (2):
(v1 - 37.5)t = v1²/(2 x 3.278)
(v1 - 37.5)t = 0.153 ----- (4)
solve 3 and 4;
(1.639t + 37.5 - 37.5)t = 0.153
1.639t² = 0.153
t² = 0.0933
t = 0.31 s
<h3>Speed of the police officer</h3>
v1 = 1.639(0.31) + 37.5 = 38 m/s = 136.8 km/h
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Answer: 
Explanation:
We know that the formula for acceleration is given by:
, where v = Final velocity
u= Initial velocity
Given : The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration.
i.e. u= 110 km/h 
[∵ 1 km= 100 meters and 1 hour = 3600 seconds]
v= 0 m/s ( At brake , final velocity becomes 0)
t=4.5 seconds
Substitute all the values in the formula , we get
Hence, the average acceleration of the car during braking is .
Aggregat 4, Aggregat 4, Aggregat 4, Aggregat 4,
