A speeder traveling at 28.8 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 2.03 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?
d1 = d2.
28.8t = 0.5*2.03*t^2
28.8t = 1.015t^2
1.015t^2 - 28.8t = 0
Use Quadratic Formula and get:
t = 28.4 s.
To find out how long it takes the policeman to catch the speeder, we can use kinematic equations of motion. The equation that relates displacement, initial velocity, final velocity, acceleration, and time is:
\(s = ut + \frac{1}{2}at^2\)
Where:
- s is the displacement
- u is the initial velocity
- t is the time taken
- a is the acceleration
Let's assume that the speeder and the policeman meet at a certain distance from the starting point. For the policeman to catch the speeder, their displacements need to be the same.
Displacement of the speeder = Displacement of the policeman
Now let's calculate the time it takes for the policeman to catch the speeder using the equation mentioned above.
For the speeder:
- Initial velocity, u = 28.8 m/s
- Final velocity, v = 0 m/s (since the speeder is caught)
- Acceleration, a = 0 m/s^2 (since the speeder maintains a constant speed)
- Displacement, s = unknown (let's call it S)
For the policeman:
- Initial velocity, u = 0 m/s (at rest)
- Acceleration, a = 2.03 m/s^2
- Displacement, s = S
Using the first equation, we can write the displacement equation for both the speeder and the policeman.
Speeder's equation:
\(S = (28.8)t + \frac{1}{2}(0)t^2\)
Simplifying, we get: \(S = 28.8t\)
Policeman's equation:
\(S = (0)t + \frac{1}{2}(2.03)t^2\)
Simplifying, we get: \(S = 1.015t^2\)
Since both displacements are equal, we can set the two equations equal to each other:
\(28.8t = 1.015t^2\)
Rearranging the equation, we get a quadratic equation:
\(1.015t^2 - 28.8t = 0\)
To solve this equation, we can use the quadratic formula:
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Where:
- a = 1.015
- b = -28.8
- c = 0
Plugging the values into the formula, we get:
\(t = \frac{-(-28.8) \pm \sqrt{(-28.8)^2 - 4(1.015)(0)}}{2(1.015)}\)
Simplifying, we get:
\(t = \frac{28.8 \pm \sqrt{829.44}}{2.03}\)
Now, we have two possible values for t: one positive and one negative. Since time cannot be negative in this context, we can ignore the negative value.
Calculating the positive value, we get:
\(t = \frac{28.8 + \sqrt{829.44}}{2.03}\)
Evaluating this expression using a calculator, we find that t is approximately 14.1 seconds.
Therefore, to the nearest tenth of a second, it takes the policeman approximately 14.1 seconds to catch the speeder.