Question

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also traveling at 20.0 m/s, and its front bumper is 24.0 m behind the truck's rear bumper. The car accelerates at a constant 0.600 m/s\(^2\), then pulls back into the truck's lane when the rear of the car is 26.0 m ahead of the front of the truck. The car is 4.5 m long, and the truck is 21.0 m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

Short answer

(a) 15.86 s, (b) 396.72 m, (c) 29.52 m/s.

Step-by-step solution

Determine Passing Distance

Calculate the total distance the car needs to travel relative to the truck. Initially, the car is 24.0 m behind, and it needs to be 26.0 m ahead after passing. Totaling these, along with the length of both the car and the truck, the car needs to move 24.0 m + 21.0 m (truck's length) + 26.0 m + 4.5 m (car's length) = 75.5 m. So the relative distance to cover is 75.5 m.

Calculate Time Required to Pass

Using the equation for uniformly accelerated motion, where the relative initial speed \(v_i = 0\) (as the car just begins accelerating to pass the truck), we use: \( s = v_i t + \frac{1}{2} a t^2 \). Substitute \( s = 75.5 \text{ m}, v_i = 0 \text{ m/s}, a = 0.600 \text{ m/s}^2 \): \( 75.5 = \frac{1}{2} \, (0.600) \, t^2 \). Solving gives: \( t^2 = 251.67 \) and \( t \approx 15.86 \text{ s} \).

Distance Traveled by the Car

To find the total distance the car travels, use the equation for distance under constant acceleration: \( s = v_i t + \frac{1}{2} a t^2 \). Here \(v_i = 20.0 \text{ m/s}\) (relative to ground), \( t = 15.86 \text{ s}, a = 0.600 \text{ m/s}^2 \): \( s = 20.0 \cdot 15.86 + \frac{1}{2} \, (0.600) \, (15.86)^2 \). Calculating we find \( s \approx 396.72 \text{ m} \).

Final Speed of the Car

To find the final speed \(v_f\), use the equation \( v_f = v_i + a t \). Here \( v_i = 20.0 \text{ m/s}, a = 0.600 \text{ m/s}^2, t = 15.86 \text{ s} \): \( v_f = 20.0 + 0.600 \cdot 15.86 \approx 29.52 \text{ m/s} \).

Key concepts

Uniform Accelerated Motion

In the context of the problem where the car must pass a truck, we encounter the concept of Uniform Accelerated Motion. Simply put, this is when an object speeds up with a constant acceleration, meaning its speed increases at a steady rate. In our scenario, the car accelerates at 0.600 m/s\(^2\), meaning every second, its speed increases by 0.600 meters per second.

This type of motion is characterized by predictable patterns, thanks to the regularity in acceleration. The car's speed changes in a uniform way, making it possible to calculate how long it will take to reach certain speeds or travel certain distances using kinematic equations.

When analyzing Uniform Accelerated Motion, it's crucial to identify:

• Initial velocity - the speed at which the object starts. For the car, despite accelerating, it initially travels at 20.0 m/s.
• Acceleration - the constant rate at which velocity changes. Here, it's the car’s increase by 0.600 m/s\(^2\).
• Time - the period over which acceleration occurs.

This consistency in acceleration allows us to use equations to solve for unknown variables, like the time required for the car to pass the truck.

Kinematic Equations

Kinematic Equations are tools to predict an object's future motion without requiring knowledge of the forces that cause that motion. These equations are essential in solving problems involving objects in Uniform Accelerated Motion, like our car overtaking the truck.

The primary equations used include:

• Displacement \( s = v_i t + \frac{1}{2} a t^2 \) - This helps find out how far an object will travel over time if it starts with an initial speed \( v_i \) and moves with constant acceleration \( a \).
• Final Velocity \( v_f = v_i + a t \) - This calculates the speed of a moving object after a certain time.

In the car-truck problem:

- The displacement equation helps us determine how long it takes the car to cover the 75.5 meters relative to the truck.
- The final velocity equation is used to find the car's speed after this time has passed, telling us how fast it is moving as it finishes the maneuver.

Using these equations allows us to efficiently solve for time, distance, and the final speed in scenarios like this and is a fundamental aspect of physics problem solving.

Physics Problem Solving

Solving physics problems is often about breaking down the situation into smaller, manageable pieces using known concepts and equations. For example, when dealing with relative motion like a car passing a truck, it's vital to analyze and synthesize different pieces of information.

Here's a common approach to physics problem solving:

• Understand the problem - Carefully read the problem statement. Identify what is given, such as initial speeds, distances, and required tasks.
• Devise a plan - Choose appropriate formulas that align with the type of motion involved, such as kinematic equations in Uniform Accelerated Motion scenarios.
• Carry out the plan - Make necessary calculations carefully. Solve for time, distance, speed, or other required variables.
• Reflect - Think about the solution and verify its accuracy. Consider if the answer makes sense in the context of the scenario.

In the car-truck example, understanding how the car accelerates and moves relative to the truck was central. By methodically applying kinematic equations step by step, we reach a comprehensive solution that aligns with the physical reality of the scenario.