Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
(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.
Time of motion of the police
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
Speed of the police officer
v1 = 1.639(0.31) + 37.5 = 38 m/s = 136.8 km/h
Learn more about velocity here: https://brainly.com/question/4931057
#SPJ1
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.