Question

A speeding motorcyclist is traveling at a constant speed of \(36.0 \mathrm{~m} / \mathrm{s}\) when he passes a police car parked on the side of the road. The radar, positioned in the police car's rear window, measures the speed of the motorcycle. At the instant the motorcycle passes the police car, the police officer starts to chase the motorcyclist with a constant acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^{2}\) a) How long will it take the police officer to catch the motorcyclist? b) What is the speed of the police car when it catches up to the motorcycle? c) How far will the police car be from its original position?

Short answer

Answer: It will take the police officer 9 seconds to catch the motorcyclist. The speed of the police car at that time will be 36.0 m/s, and the police car will be 324 meters from its original position.

Step-by-step solution

Write down the equations for motion

Using the equations for uniform linear motion for the motorcyclist (M) and accelerated linear motion for the police car (P): Motorcyclist: \(x_M = v_M t\) Police car: \(x_P = v_P0 * t + \frac{1}{2} a_P t^2\)

Set the distances equal and solve for time

When the police car catches the motorcyclist, their distances from the starting point will be equal: \(x_M = x_P\). Replace the equations from step 1 and set them equal to each other: \(v_M t = v_P0 * t + \frac{1}{2} a_P t^2\) Since the police car starts at rest, \(v_P0 = 0\). Then, plug the given values (\(v_M = 36.0 \mathrm{~m}/\mathrm{s}\) and \(a_P = 4.0 \mathrm{~m}/\mathrm{s}^{2}\)) into the equation: \(36.0t = \frac{1}{2}(4.0)t^2\)

Solve for t

To make the equation easier to solve, divide both sides by \(t\) and 2: \(18.0 = 2.0t\) Now, solve for \(t\): \(t = \frac{18.0}{2.0} = 9.0\text{ s}\) The police officer will catch the motorcyclist in 9 seconds. b) What is the speed of the police car when it catches up to the motorcycle?

Find the final speed of the police car

To find the final speed of the police car, we can use the equation for the final velocity in uniform accelerated motion: \(v_P = v_P0 + a_Pt\) Plug in the given values and the time calculated in part (a) for t: \(v_P = 0 + (4.0\mathrm{~m}/\mathrm{s}^{2})(9\text{ s}) = 36.0\mathrm{~m}/\mathrm{s}\) The speed of the police car when it catches up to the motorcyclist is 36.0 m/s. c) How far will the police car be from its original position?

Calculate the distance the police car covers

To find the distance the police car covers, we can use either the equation for the motorcyclist or the police car since their distances are equal when the police car catches the motorcyclist. We'll use the motorcyclist's equation: \(x_M = v_M t\) Plug in the given value for \(v_M\) and the time calculated in part (a) for t: \(x_M = (36.0\mathrm{~m}/\mathrm{s})(9\text{ s}) = 324\mathrm{~m}\) The police car will be 324 meters from its original position when it catches up to the motorcyclist.

Key concepts

Physics Kinematics

Physics kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause this motion. Let's explore this through our motorcycle and police car scenario.

In kinematics, we describe an object's movement using terms such as distance, displacement, speed, velocity, and acceleration. Speed is a scalar quantity representing how fast an object is moving, while velocity is a vector quantity that describes both the speed and direction of motion. Acceleration is the rate of change of velocity over time.

For our motorcyclist moving at a constant speed, kinematic equations are relatively simple because the velocity does not change over time. However, the police car, which accelerates to catch up with the motorcyclist, requires more complex kinematic analysis. Here, we use equations that incorporate acceleration to model the police car's motion.

Equations of Motion

Equations of motion are mathematical formulas that describe the relationship between displacement, velocity, acceleration, and time. They are fundamental tools in kinematics, allowing us to predict future motion based on current and past motion parameters.

For uniformly accelerated motion, the three key equations are:

• The first equation relates velocity to time: \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time elapsed.
• The second equation connects displacement and time: \(s = ut + \frac{1}{2}at^2\), with \(s\) representing displacement.
• The third equation links displacement with velocity: \(v^2 = u^2 + 2as\).

In the given problem, we used the second equation to determine the time it takes for the police car to catch the motorcyclist, as well as to calculate the final speed of the police car and the displacement covered during this time.

Motion with Constant Acceleration

In our exercise, the police car exhibits motion with constant acceleration. This type of motion is characterized by a steady increase in velocity, which happens at a uniform rate. It is an essential concept in kinematics as it often applies to real-world situations, such as vehicles speeding up on roads.

The uniform acceleration of the police car allows us to use specific kinematic equations for calculation. Since the initial velocity of the police car is zero (\(v_{P0} = 0\text{ m/s}\)), our calculations become simpler. The formula we use to calculate the time taken to catch up with the motorcyclist essentially equates the distance covered by both vehicles, considering the motorcyclist moves at a constant speed and the police car speeds up from rest with a specified acceleration.

Understanding motion with constant acceleration is crucial for solving many physics problems and is particularly applicable in designing transportation systems, predicting the motion of vehicles, and in engineering tasks related to vehicle safety and performance.