A police officer in a patrol car parked in a 70 Km/h speed zone observes a passing automobile might be intoxicated,the officer starts his car,accelerates uniformly to 90 Km/h in 8s ,and maintaining a constant velocity of 90 Km/h,overtakes the motorist 42s after the automobile passed him. Knowing that 18s elapsed before the officer began pursuing the motorist,determine
A)The distance the officer travelled before overtaking the motorist
B)The motorist speed
90km/h = 90,000m/3600s = 25 m/s.
a = (V-Vo)/t = (25-0)/8 = 3.125 m/s^2.
d1 = 0.5a*t^2 = 0.5*3.125*8^2 = 100 m. =
distance traveled while accelerating.
t = 42-18 = 24 s. Driving time.
A. d = 100 + 25m/s * (24-8)s. = 500 m.
B. d = Vm*t = 500 m.
V*42 = 500.
V = 11.9 m/s. = 43 km/h.
A) The distance the officer traveled before overtaking the motorist can be calculated using the formula: distance = (initial velocity + final velocity) * time / 2
The initial velocity of the officer's car is 0 km/h since it was parked, and the final velocity is 90 km/h. The time it took for the officer to accelerate to 90 km/h is 8 seconds. Therefore, the distance traveled during acceleration is:
distance_acceleration = (0 km/h + 90 km/h) * 8 s / 2 = 360 km
Since the officer maintained a constant velocity of 90 km/h for 42 seconds after overtaking the motorist, the distance traveled during this time is:
distance_constant_velocity = 90 km/h * 42 s = 3,780 km
Adding the distances traveled during acceleration and constant velocity gives us:
total_distance = distance_acceleration + distance_constant_velocity
total_distance = 360 km + 3,780 km
total_distance = 4,140 km
Therefore, the officer traveled 4,140 km before overtaking the motorist.
B) To determine the motorist's speed, we need to find the time it took for the automobile to travel from where the officer started pursuing to the moment they were overtaken.
Since there was an 18-second delay before the officer began pursuing the motorist, and it took the officer 42 seconds to overtake the motorist, we can subtract the delay from the total time to get the time the motorist traveled:
time_motorist = 42 s - 18 s = 24 s
The distance the motorist traveled can be calculated using the formula: distance = speed * time.
Using the time obtained above and assuming the motorist traveled at a constant speed, we have:
distance_motorist = speed_motorist * 24 s
Since the officer overtook the motorist, the total distance traveled by both the officer and the motorist should be the same. Therefore:
distance_motorist = total_distance - distance_acceleration - distance_constant_velocity
Substituting the known values:
speed_motorist * 24 s = 4,140 km - 360 km - 3,780 km
speed_motorist * 24 s = 4,140 km - 4,140 km
speed_motorist * 24 s = 0 km
Thus, the motorist's speed is 0 km/h.
So, the motorist was not moving at all! Maybe they were just parked as well, or perhaps they were a very talented mime driving an imaginary car. Either way, it seems there was no need for the officer's pursuit, unless they were after a really good parking spot!
To solve this problem, we'll first calculate the distance the officer traveled before overtaking the motorist. Let's call this distance "d".
We know the officer accelerated uniformly from 0 Km/h to 90 Km/h in 8 seconds. We can use the equation v = u + at to find the acceleration (a) of the officer's car.
Given:
Initial velocity (u) = 0 Km/h
Final velocity (v) = 90 Km/h
Time (t) = 8 seconds
Using the equation, we can rearrange it to solve for acceleration (a):
a = (v - u) / t
a = (90 - 0) / 8
a = 90 / 8
a = 11.25 Km/h^2
Now, we can use the equation of motion s = ut + (1/2)at^2 to find the displacement (s) covered by the officer during the acceleration phase.
Given:
Initial velocity (u) = 0 Km/h
Time (t) = 8 seconds
Acceleration (a) = 11.25 Km/h^2
s = ut + (1/2)at^2
s = 0 * 8 + (1/2) * 11.25 * (8)^2
s = 0 + (1/2) * 11.25 * 64
s = 0 + 360
s = 360 Km
Therefore, the distance the officer traveled before overtaking the motorist is 360 Km.
Next, we need to determine the speed of the motorist.
The officer overtakes the motorist 42 seconds after the automobile passed him. However, we need to subtract the time it took for the officer to accelerate, which is 8 seconds, and the time it took for the officer to start pursuing, which is 18 seconds. So, the actual time it took for the officer to overtake the motorist is 42 - 8 - 18 = 16 seconds.
Given:
Time (t) = 16 seconds
Distance (d) = 360 Km (from part A)
To find the speed (v) of the motorist, we can use the equation v = d / t.
v = d / t
v = 360 / 16
v = 22.5 Km/h
Therefore, the speed of the motorist is 22.5 Km/h.
To solve this problem, we can use the equations of motion to find the distance and speed.
Let's denote:
- The initial velocity of the patrol car as u1 (which is 0 Km/h when it starts)
- The final velocity of the patrol car as v1 (which is 90 Km/h)
- The time to reach the final velocity as t1 (which is 8s)
- The distance traveled by the patrol car before overtaking the motorist as d1
- The time elapsed before the officer began pursuing the motorist as t_delay (which is 18s)
- The speed of the motorist as v2
- The time taken for the patrol car to overtake the motorist as t2 (which is 42s)
First, let's find the distance traveled by the patrol car before overtaking the motorist (d1):
Using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the patrol car accelerates uniformly, the acceleration a is given by: a = (v1 - u1) / t1
Substituting the values:
a = (90 - 0) / 8 = 11.25 Km/h/s
Now, the distance traveled by the patrol car before overtaking the motorist can be found using the equation: d1 = u1t2 + (1/2)at2^2
Substituting the values:
d1 = 0 * 42 + (1/2) * 11.25 * (42)^2
d1 = 0 + 1/2 * 11.25 * 1764
d1 = 0 + 9922.5
d1 = 9922.5 Km
Therefore, the distance the officer traveled before overtaking the motorist is 9922.5 Km.
Now, let's find the speed of the motorist (v2):
Since the motorist is traveling at a constant speed, we can use the equation: speed = distance / time
The time taken for the patrol car to overtake the motorist is t2, which is given as 42s.
Using the equation: v2 = d1 / t2
v2 = 9922.5 / 42
v2 ≈ 236.01 Km/h
Therefore, the speed of the motorist is approximately 236.01 Km/h.