A speeding motorist traveling 105 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 11.0 km/h/s (note the mixed units).
(a) How much time will it take for the police officer to reach the speeder, assuming that the speeder maintains a constant speed?
To determine the time it will take for the police officer to reach the speeder, we need to use the equation of motion.
Let's first convert the speeds given into meters per second (m/s):
Speed of the motorist = 105 km/h
1 km = 1000 m
1 hour = 3600 seconds
So, the speed of the motorist = (105 km/h) * (1000 m/km) / (3600 s/h) ≈ 29.17 m/s
Acceleration of the police officer = 11.0 km/h/s
So, the acceleration of the police officer = (11.0 km/h) * (1000 m/km) / (3600 s/h) ≈ 3.06 m/s²
Now, let's use the equation of motion to find the time it takes for the police officer to reach the speeder, assuming the speeder maintains a constant speed.
The equation of motion for constant acceleration is: s = ut + 0.5 * a * t²
Where:
s = distance traveled
u = initial velocity
a = acceleration
t = time
In this case, the distance traveled by both the speeder and the police officer when the officer catches up to the speeder will be the same, as the speeder is maintaining a constant speed.
The speeder's distance traveled = speed of the motorist * t
The police officer's distance traveled = initial velocity of the police officer * t + 0.5 * acceleration of the police officer * t²
Since both distances are the same, we can equate them:
Speed of the motorist * t = initial velocity of the police officer * t + 0.5 * acceleration of the police officer * t²
Simplifying the equation, we get:
29.17t = 0 + 0.5 * 3.06 * t²
0.5 * 3.06 * t² - 29.17t = 0
Now we can solve this quadratic equation for t using the quadratic formula.
t = (-b ± sqrt(b² - 4ac)) / (2a)
For this equation, a = 0.5 * 3.06, b = -29.17, and c = 0.
Plugging these values into the quadratic formula, we get:
t = (-(-29.17) ± sqrt((-29.17)² - 4 * 0.5 * 3.06 * 0)) / (2 * 0.5 * 3.06)
t = (29.17 ± sqrt(853.3489)) / 3.06
t ≈ (29.17 ± 29.20) / 3.06
So, the two solutions for t are:
t1 ≈ (29.17 + 29.20) / 3.06 ≈ 19.05 seconds
t2 ≈ (29.17 - 29.20) / 3.06 ≈ -0.01 seconds
Since time cannot be negative, the solution t2 is not valid.
Therefore, it will take approximately 19.05 seconds for the police officer to reach the speeder, assuming the speeder maintains a constant speed.
To solve this problem, we need to find the time it will take for the police officer to catch up with the speeding motorist.
Let's convert all the units to a common unit to make calculations easier. Since the officer's acceleration is given in km/h/s, we need to convert it to m/s².
1 km = 1000 m
1 h = 3600 s
So, the officer's acceleration is:
11.0 km/h/s = (11.0 km/h) * (1000 m/1 km) * (1 h/3600 s) = (11.0 * 1000) / 3600 m/s² ≈ 3.06 m/s²
Now, let's convert the speed of the motorist to m/s:
105 km/h = (105 km) * (1000 m/1 km) * (1 h/3600 s) = (105 * 1000) / 3600 m/s ≈ 29.17 m/s
To find the time it will take for the officer to catch up with the motorist, we'll use the second equation of motion:
v = u + at
Where:
v = final velocity (officer's velocity)
u = initial velocity (officer's initial velocity)
a = acceleration
t = time
Since the motorist maintains a constant speed, their final velocity is equal to the officer's velocity. Therefore, we have:
v = 29.17 m/s
u = 0 m/s (officer's initial velocity at rest)
a = 3.06 m/s²
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (29.17 m/s - 0 m/s) / 3.06 m/s² = 9.53 s
Therefore, it will take approximately 9.53 seconds for the police officer to reach the speeder, assuming the speeder maintains a constant speed.