Question

Two train cars are on a straight, horizontal track. One car starts at rest and is put in motion with a constant acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). This car moves toward a second car that is \(30.0 \mathrm{~m}\) away. The second car is moving away from the first car and is traveling at a constant speed of \(4.00 \mathrm{~m} / \mathrm{s}\). a) Where will the cars collide? b) How long will it take for the cars to collide?

Short answer

Answer: The cars will collide at a position approximately 61.1 meters away from the starting point of the first car, and it will take approximately 7.85 seconds for the cars to collide.

Step-by-step solution

Write the equations of motion for both cars

We have information about both cars, so we can write the equations of motion for each one of them. For the first car, it starts at rest and accelerates at a constant rate. Thus, we can use the equation: \(x_1(t) = \frac{1}{2}at^2\) where \(x_1(t)\) is the position of the first car at time \(t\), \(a\) is the acceleration, and \(t\) is the time. For the second car, it moves at a constant speed. Thus, we can use the equation: \(x_2(t) = v_0t + x_0\) where \(x_2(t)\) is the position of the second car at time \(t\), \(v_0\) is the constant speed, and \(x_0\) is the initial position of the second car.

Plug in the given values for acceleration, speed, and initial distance

We are given the values of \(a=2.00\,\text{m}/\text{s}^2\), \(v_0=4.00\:\text{m}/\text{s}\), and \(x_0=30.0\,\text{m}\). Plugging these values in the two equations of motion, we get: \(x_1(t) = (1.00 \mathrm{~m} / \mathrm{s}^2) t^2\) \(x_2(t) = (4.00 \mathrm{~m} / \mathrm{s}) t + 30.0\,\text{m}\)

Set the position equations equal to each other to find the time of collision

Since we want to find when the two cars collide, we need to find the time \(t\) when their positions are equal. We can set the two position equations equal to each other and solve for the time: \((1.00 \mathrm{~m} / \mathrm{s}^2) t^2 = (4.00 \mathrm{~m} / \mathrm{s}) t + 30.0\,\text{m}\) Rearrange the equation to have a quadratic form: \(t^2 - 4\mathrm{~m} / \mathrm{s} t - 30.0\,\text{m} = 0\)

Solve the quadratic equation for time

Use any preferred method (factoring, quadratic formula, etc.) to find the positive solution to the quadratic equation. In this case, we'll use the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=1\), \(b=-4\:\text{m}/\text{s}\), and \(c=-30\,\text{m}\). Using the quadratic formula, we obtain two solutions: \(t_1 \approx -3.85\,\text{s}\) (discard this solution since time must be positive) \(t_2 \approx 7.85\,\text{s}\) (accept this solution) Thus, it will take approximately \(7.85\,\text{s}\) for the two cars to collide.

Determine the position of collision

We analyze the equations of motion again. Now that we have the time it takes for the cars to collide, we can find the position of the collision. We can plug the time \(t_2\) into either of the position equations (let's use \(x_1(t)\)): \(x_1(7.85\,\text{s}) = (1.00 \mathrm{~m} / \mathrm{s}^2) (7.85\,\text{s})^2 \approx 61.1\,\text{m}\) Therefore, the cars will collide at a position approximately \(61.1\,\text{m}\) from the starting point of the first car. So, the answers to the questions are: a) The cars will collide at a position approximately \(61.1\,\text{m}\) away from the starting point of the first car. b) It will take approximately \(7.85\,\text{s}\) for the cars to collide.

Key concepts

Equations of Motion

The equations of motion are mathematical formulas that describe the movement of objects under the influence of forces. In kinematics, these equations help us predict the future position, velocity, and acceleration of an object when it is moving under uniform acceleration. The basic equations include:

• \tThe equation for the position as a function of time: \( x(t) = x_0 + v_0t + \frac{1}{2}at^2 \)
• \tThe velocity-time relationship: \( v(t) = v_0 + at \)
• \tThe equation for displacement based on velocity and acceleration: \( x = v_0t + \frac{1}{2}at^2 \)

In the case of the problem involving the two train cars, we applied these equations to analyze their motion and predict where and when they would collide.

Constant Acceleration

Constant acceleration occurs when an object's velocity changes at a consistent rate over time. This situation is common in physics problems and is a key assumption that simplifies the equations of motion. In the train cars example, the first car starts from rest and accelerates at a constant \(2.00\,\text{{m/s}}^2\), which allows us to use a simplified equation \(x_1(t) = \frac{1}{2}at^2\) to describe its motion.

Constant Speed

In contrast to constant acceleration, constant speed implies that an object moves at a steady pace, covering equal distances in equal time intervals. There is no acceleration because the velocity doesn't change. For the second car in the problem that moves away at a constant speed of \(4.00\,\text{{m/s}}\), the position as a function of time is given by \(x_2(t) = v_0t + x_0\), reflecting its uniform motion.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, generally in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. To find the roots or solutions to the equation, we can use various methods such as factoring, completing the square, or applying the quadratic formula \(t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). In kinematics problems, such as the collision of the two cars, a quadratic equation may arise when solving for variables like time or position, and choosing the physically meaningful solution is crucial.

Position-Time Relationship

The position-time relationship is fundamental in kinematics for describing how an object's position depends on time. This relationship is represented by a mathematical function that varies depending on the type of motion—whether it's uniform motion or accelerated motion. In our train car scenario, setting the position-time equations equal for both cars determines the point of collision. By finding the time when their positions match, we can calculate the exact moment and position of the collision using the derived function.