A police car is traveling at a velocity of 15.0 m/s due north, when a car zooms by at a constant velocity of 44.0 m/s due north. After a reaction time 0.700 s the policeman begins to pursue the speeder with an acceleration of 5.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?
d1 = d2
44(t+0.7) = Vo*t + 0.5a*t^2
44t + 30.8 = 15t + 2.5t^2
2.5t^2 + 15t-44t-30.8 = 0
2.5t^2 -29t - 30.8 = 0
Use Quadratic Formula and get:
t = 12.6 s.
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To find out how long it takes for the police car to catch up with the speeder, we need to determine the distance the speeder travels during the policeman's reaction time, and then calculate the time it takes for the police car to cover that distance.
Let's start by finding out the distance the speeder travels during the reaction time of 0.700 s.
We know that the speeder's velocity is 44.0 m/s due north, and we need to find the distance traveled. The formula to calculate distance is:
distance = velocity × time
So, during the reaction time of 0.700 s, the distance traveled by the speeder is:
distance = velocity × time
distance = 44.0 m/s × 0.700 s
distance = 30.8 m
The speeder travels 30.8 meters during the policeman's reaction time.
Next, we need to calculate how long it takes for the police car to cover this distance with an acceleration of 5.00 m/s^2.
To find the time it takes for the police car to cover a certain distance with a given acceleration, we can use the kinematic equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
In this case, the distance is 30.8 meters, the initial velocity is 15.0 m/s (as given in the question), and the acceleration is 5.00 m/s^2. We need to find the time.
Plugging the values into the equation, we get:
30.8 m = 15.0 m/s × t + 0.5 × 5.00 m/s^2 × t^2
Simplifying the equation:
30.8 m = 15.0 m/s × t + 2.5 m/s^2 × t^2
Rearranging the equation:
2.5 m/s^2 × t^2 + 15.0 m/s × t - 30.8 m = 0
Now, we can solve this quadratic equation to find the value of t.
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Where a = 2.5, b = 15.0, and c = -30.8.
Plugging in the values, we get:
t = (-15.0 ± sqrt(15.0^2 - 4 × 2.5 × -30.8)) / (2 × 2.5)
Simplifying further:
t = (-15.0 ± sqrt(225 + 310)) / 5
t = (-15.0 ± sqrt(535)) / 5
Using a calculator, we find:
t ≈ 2.68 s or t ≈ -1.48 s
Since time cannot be negative in this context, we discard the negative value.
Therefore, the time it takes for the police car to catch up with the speeder, including the reaction time, is approximately:
Reaction time + time to cover distance = 0.700 s + 2.68 s = 3.38 s
So, it takes approximately 3.38 seconds for the police car to catch up with the speeder.