(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
Learn more about velocity here: brainly.com/question/4931057
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