A police car is traveling at a constant velocity 18 m/s north, when a car zooms by at a constant velocity 42 m/s due north. After a reaction time of 0.8 s the police car pursue the speeder with an acceleration of 5 m/s2. How long (in s) does it take for the police car to catch the speeder? Include reaction time of police car.
Well, it seems like the police car is going to have a little high-speed chase! Let's break it down.
First, let's consider the time it takes for the police car to react, which is 0.8 seconds. During this time, both the police car and the speeder are still moving at their original velocities, since there is no acceleration yet.
Now, let's focus on the situation after the police car starts accelerating. We can think of this as a race between the police car and the speeder, with the police car trying to catch up.
The speeder is moving at a constant velocity of 42 m/s north, while the police car is initially moving at 18 m/s north. However, the police car is accelerating at 5 m/s².
So, the police car is going to catch up by increasing its velocity faster than the speeder is maintaining theirs. The relative speed between the two is the difference between their velocities: 42 m/s - 18 m/s = 24 m/s.
Now, we can use the equation of motion: v = u + at, where:
v = final velocity (relative speed in this case)
u = initial velocity (difference in velocities between the two)
a = acceleration (5 m/s²)
t = time
Substituting the values, we get: 24 m/s = 0 + 5 m/s² * t
Now, let's find the time it takes for the police car to catch the speeder.
24 m/s = 5 m/s² * t
t = 24 m/s / 5 m/s²
t = 4.8 seconds
So, even with a reaction time of 0.8 seconds, it will take the police car 4.8 seconds to catch the speeder. I hope that gives you a clownish perspective on the situation!
To find out how long it takes for the police car to catch the speeder, we need to calculate the time it takes for the police car to match the speeder's velocity and overtake it.
1. Calculate the distance covered by the speeder during the reaction time of the police car:
Distance = Velocity × Time
Distance = 42 m/s × 0.8 s
Distance = 33.6 m
2. Now, we need to calculate the relative velocity between the police car and the speeder:
Relative velocity = Speeder's velocity - Police car's velocity
Relative velocity = 42 m/s - 18 m/s
Relative velocity = 24 m/s (since both velocities are in the same direction)
3. Next, we need to find the time it takes for the police car to catch up to the speeder. We can use the equation of motion:
Distance = Initial velocity × Time + 0.5 × Acceleration × Time^2
The initial velocity of the police car is 0 m/s (since it starts from rest after the reaction time).
Acceleration = 5 m/s^2
Distance = Relative velocity × Time + 0.5 × Acceleration × Time^2
Plugging in the values and rearranging the equation to solve for time:
33.6 m + 0.5 × 5 m/s^2 × Time^2 = 24 m/s × Time
Simplifying the equation:
33.6 m = 24 m/s × Time - 2.5 m/s^2 × Time^2
Rearranging the equation:
2.5 m/s^2 × Time^2 - 24 m/s × Time + 33.6 m = 0
This is a quadratic equation that can be solved to find the value of Time.
4. Solve the quadratic equation to find the time:
Using the quadratic formula:
Time = [-(-24 m/s) ± sqrt((-24 m/s)^2 - 4 × 2.5 m/s^2 × 33.6 m)] / (2 × 2.5 m/s^2)
Solving for Time gives two possible values. However, we only need the positive value for time:
Time = [24 m/s ± sqrt(576 m^2/s^2 - 4 × 2.5 m/s^2 × 33.6 m)] / 5 m/s^2
Time = [24 m/s ± sqrt(576 m^2/s^2 - 336 m^2/s^2)] / 5 m/s^2
Time = [24 m/s ± sqrt(240 m^2/s^2)] / 5 m/s^2
Time = [24 m/s ± 15.49 m/s] / 5 m/s^2
Time = (24 m/s + 15.49 m/s) / 5 m/s^2 (Discarding the negative solution)
Time = 39.49 m/s / 5 m/s^2
Time ≈ 7.898 s
Therefore, it takes approximately 7.898 seconds for the police car to catch the speeder, including the reaction time.
To determine how long it takes for the police car to catch the speeder, we need to analyze the motion of both vehicles.
Let's break down the information given:
1. Initial velocity of the police car (vP): 18 m/s north (positive velocity implies the north direction).
2. Initial velocity of the speeder car (vS): 42 m/s north.
3. Reaction time of the police car (tR): 0.8 s.
4. Acceleration of the police car (aP): 5 m/s^2 (constant acceleration in pursuit).
First, we need to determine the distance covered by both vehicles during the police car's reaction time.
For the police car:
Distance covered by the police car during reaction time (dP) = (vP) * (tR)
dP = 18 m/s * 0.8 s
For the speeder car:
Distance covered by the speeder car during reaction time (dS) = (vS) * (tR)
dS = 42 m/s * 0.8 s
Now, let's determine the time it takes for the police car to catch up with the speeder car, starting from the end of the reaction time.
The relative velocity between the police car and the speeder car is obtained by subtracting their initial velocities:
Relative velocity (vR) = vS - vP = 42 m/s - 18 m/s
Using the formula:
Distance (d) = Initial velocity (v) * Time (t) + (1/2) * Acceleration (a) * (Time^2)
We can determine the time (t) for the police car to catch the speeder car by solving the equation:
dP + dS + vR * t + (1/2) * aP * (t^2) = 0
Substituting the known values:
18 m/s * 0.8 s + 42 m/s * 0.8 s + (42 m/s - 18 m/s) * t + (1/2) * 5 m/s^2 * (t^2) = 0
Simplifying the equation, we get a quadratic equation which can be solved to find the value of t.
d1 = V*t = 42m/s * 0.8s = 33.6 m. Head start for the speeder.
d2 = d1 + 33.6 m.
20*t + 0.5*a*t^2 = 42*t + 33.6
20t+ 0.5*5*t^2 = 42t + 33.6
2.5t^2 -22t = 33.6
2.5t^2 - 22t - 33.6 = 0
Use Quadratic Formula.
t = 10.13 s.