An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
im still a bit confused, i still don't know the answer to the question
To solve this problem, we can use the formula for distance:
Distance = Initial velocity * Time + 0.5 * Acceleration * Time^2
We are given:
- Initial velocity of the police car (u1) = 95 km/h
- Velocity of the speeder (u2) = (95 + 14) km/h = 109 km/h
- Acceleration of the police car (a) = 2.30 m/s^2
- Time when the speeder passes the police car (t1) = 2.50 s
First, let's convert the velocities to meters per second (m/s) and the time to seconds (s):
1 km/h = (1000 m)/(3600 s) = 0.2778 m/s
u1 = 95 km/h * 0.2778 m/s = 26.39 m/s
u2 = 109 km/h * 0.2778 m/s = 30.29 m/s
t1 = 2.50 s
Now, we can find the distance traveled by the speeder during t1:
Distance1 = u2 * t1
= 30.29 m/s * 2.50 s
= 75.725 m
Next, let's find the time it takes for the police car to catch up with the speeder. We can use the equation:
Distance1 + Distance2 = u1 * t2 + 0.5 * a * t2^2
Since the speeder is moving at a constant speed, Distance2 is equal to u2 * t2:
Distance1 + u2 * t2 = u1 * t2 + 0.5 * a * t2^2
Rearranging the equation, we get:
0.5 * a * t2^2 + (u1 - u2) * t2 - Distance1 = 0
Substituting the given values:
0.5 * 2.30 * t2^2 + (26.39 - 30.29) * t2 - 75.725 = 0
Now we can solve this quadratic equation for t2 using the quadratic formula:
t2 = (-b ± √(b^2 - 4ac)) / (2a)
where a = 0.5 * 2.30, b = (26.39 - 30.29), and c = -75.725.
Plugging in these values, we get:
t2 = [-(26.39 - 30.29) ± √((26.39 - 30.29)^2 - 4 * 0.5 * 2.30 * (-75.725))] / (2 * 0.5 * 2.30)
Simplifying further:
t2 = [3.90 ± √(17.0403 + 696.955)] / (2.3)
= [3.90 ± √(713.9953)] / (2.3)
Calculating the square root:
t2 = [3.90 ± 26.737] / (2.3)
Now we have two values for t2:
t2_1 = (3.90 + 26.737) / 2.3
= 30.64 / 2.3
= 13.35 s
t2_2 = (3.90 - 26.737) / 2.3
= -22.84 / 2.3
= -9.94 s (Discard negative value)
Therefore, it takes approximately 13.35 seconds for the police car to overtake the speeder after the speeder passes.
To solve this problem, we can first calculate the distance between the police car and the speeder when the police officer starts accelerating. Then, we can find the time it takes for the police car to catch up to the speeder.
Let's break down the problem into steps:
Step 1: Calculate the distance between the police car and the speeder when the police officer starts accelerating.
The speeder is traveling at a constant speed of 14 km/h. In 2.50 seconds, the speeder covers a distance of:
Distance = Speed × Time
Distance = (14 km/h) × (2.50 s)
Distance = (14 km/h) × (2.50 s) × (1/3600 h/s) [Converting km/h to m/s]
Distance = (14 m/s) × (2.50 s)
Distance = 35 m
So, when the police officer starts accelerating, the speeder is 35 meters ahead.
Step 2: Calculate the time it takes for the police car to catch up to the speeder.
Now, we need to find how long it takes for the police car to cover the 35-meter distance and catch up to the speeder. We'll use the equations of motion to solve this.
We can use the equation of motion:
Distance = Initial Velocity × Time + (1/2) × Acceleration × Time^2
In this case, the initial velocity of the police car is 95 km/h, which we need to convert to m/s:
Initial Velocity = 95 km/h × (1/3600 h/s)
Initial Velocity = (95 m/s) × (1/3600 s)
Initial Velocity = 26.39 m/s
Using this information, we can plug in the values into the equation of motion and solve for time.
35 m = (26.39 m/s) × t + (1/2) × (2.30 m/s^2) × t^2
35 m = 26.39 m/s × t + 1.15 m/s^2 × t^2
0 = 1.15t^2 + 26.39t - 35
We solve this quadratic equation to find the time it takes for the police car to catch up to the speeder.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / 2a, we can substitute in the values:
t = [-(26.39) ± √((26.39)^2 - 4(1.15)(-35))] / (2 × 1.15)
Solving the equation gives us two possible solutions, but we discard the negative root since time cannot be negative:
t ≈ 1.19 s
Therefore, it will take approximately 1.19 seconds for the police car to overtake the speeder after the speeder passes.
Please note that in reality, the exact time may vary slightly due to factors such as the reaction time of the police officer.
The time of police car motion is t,
the time of the speeder motion is (t+2.5).
The distance covered by
- the police car is v(p) •t+at^2/2,
- by speeder is v(s)•(t+2.5).
v(p) •t + at^2/2 = v(s)•(t+2.5).
v(p) = 95 km/h = 26.4 m/s
v(s) = 140 km/h = 38.9 m/s
v(p) •t + at^2/2 = v(s)• t + v(s)• 2.5,
at^2/2 + [v(p) – v(s)] •t -2.5•v(s) = 0,
t^2 –(2/a)•[v(s) – v(p)] •t – 2.5•2/a = 0,
t^2 – 10.9t - 84.6 = 0,
t = 5.24 s.
The total time is 5.24+2.5 = 7.74 s.