Answer:
a. 18.34 s b. 327.92 m
Step-by-step explanation:
a. How long before the police car catches the speeder who continued traveling at 40 miles/hour
The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.
So, a = (v - u)/t = (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².
The distance moved by the police car in 10 s is gotten from
s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.
s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²
s = 0 m + 1/2 × 2.458 m/s² × 100 s²
s = 122.9 m
The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h
The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t
The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' = 17.88t' m where t' = time taken for police to catch up with speeder.
Since both distances are the same,
S' = S
17.88t' = 122.9 + 24.58t
Also, the time taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t
t' = 10 + t
So, substituting t' into the equation, we have
17.88t' = 122.9 + 24.58t
17.88(10 + t) = 122.9 + 24.58t
178.8 + 17.88t = 122.9 + 24.58t
17.88t - 24.58t = 122.9 - 178.8
-6.7t = -55.9
t = -55.9/-6.7
t = 8.34 s
So, t' = 10 + t
t' = 10 + 8.34
t' = 18.34 s
So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour
b. how far before the police car catches the speeder who continued traveling at 40 miles/hour
Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m
