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 and moving 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: a) The cars will collide at a position of \(25.0 \mathrm{~m}\) from Car 1's starting position. b) It will take \(5.0 \mathrm{~s}\) for the cars to collide.

Step-by-step solution

Equations of motion for Car 1

For Car 1, which starts from rest and accelerates at a constant rate of \(2.0 \mathrm{~m}/\mathrm{s}^2\), we can write the equation of motion as: \( x_1(t) = \frac{1}{2}at^2 \), where \(x_1(t)\) is the position of Car 1 as a function of time \(t\), \(a = 2.0 \mathrm{~m}/\mathrm{s}^2\) is the acceleration, and \(t\) is time.

Equations of motion for Car 2

For Car 2, moving at a constant speed of \(4.0 \mathrm{~m}/\mathrm{s}\) and initially \(30.0 \mathrm{~m}\) away from Car 1, we can write the equation of motion as: \( x_2(t) = 30.0 \mathrm{~m} + vt \), where \(x_2(t)\) is the position of Car 2 as a function of time \(t\), \(v = 4.00 \mathrm{~m}/\mathrm{s}\) is the constant speed, and \(t\) is time.

Find the collision point

The cars will collide when their positions are equal, so we can set the two position functions equal to each other: \( \frac{1}{2}at^2 = 30.0 \mathrm{~m} + vt \) Plugging the acceleration of Car 1 and the speed of Car 2 into the equation, we get: \( \frac{1}{2}(2.0 \mathrm{~m}/\mathrm{s}^2)t^2 = 30.0 \mathrm{~m} + (4.00 \mathrm{~m}/\mathrm{s})t \)

Solve for the time it takes for the cars to collide

To solve for \(t\), we can simplify the given equation: \( t^2 = 30.0 \mathrm{~m} + 4.00 \mathrm{~m}/\mathrm{s}t \) Now, combine the terms involving \(t\) and factor out \(t\) from the resulting expression to find the time: \( t^2 - 4.00\mathrm{~m}/\mathrm{s}t = 30.0 \mathrm{~m} \) \( t(t - 4.00\mathrm{~m}/\mathrm{s}) = 30.0 \mathrm{~m} \) Because \(t\geq0\), we get: \( t = 4.00\mathrm{~m}/\mathrm{s} - 30.0 \mathrm{~m} = 5.0 s \)

Calculate the collision point

Now that we've found the time it takes for the cars to collide, we can use either position equation to find the position of the collision. We'll use the equation for Car 1: \( x_1(t) = \frac{1}{2}(2.0 \mathrm{~m}/\mathrm{s}^2)(5.0 \mathrm{~s})^2 \) \( x_1(5.0 \mathrm{~s}) = 25.0 \mathrm{~m} \) Therefore, the cars will collide at a distance of \(25.0 \mathrm{~m}\) from Car 1's starting position. a) The cars will collide at a position of \(25.0 \mathrm{~m}\) from Car 1's starting position. b) It will take \(5.0 \mathrm{~s}\) for the cars to collide.

Key concepts

Constant acceleration

Constant acceleration refers to a situation where an object speeds up or slows down at a consistent rate over time. This is a common scenario in physics, especially in kinematics. When an object has constant acceleration, its velocity changes by the same amount each second.
An example of this is Car 1 in our exercise. Car 1 starts from rest, meaning its initial speed is zero. As it begins to move, it accelerates at a constant rate of \(2.00 \, \mathrm{m/s^2}\). This means that with each passing second, its speed increases by \(2.00 \, \mathrm{m/s}\).
Constant acceleration is important because it allows us to use specific equations to predict motion. These equations allow us to calculate how far something has moved over a period of time, how fast it is going at any point, and how long it takes to reach a certain speed or location.

• For instance, if you want to calculate how far Car 1 moves in a certain time, you can use the equation: \(x = \frac{1}{2}at^2\).
• This setup means distances traveled are proportional to the square of the time elapsed—a key characteristic of objects under constant acceleration.

Equations of motion

Equations of motion are mathematical formulas used to describe the behavior of moving objects. They are invaluable for solving problems where you need to predict where an object will be in the future or where it was in the past, based on its speed, initial position, and any acceleration.
In the problem at hand, we have two train cars with distinct motion characteristics. Car 1 has an equation that reflects its constant acceleration, expressed as \(x_1(t) = \frac{1}{2}at^2\), where \(a\) is the acceleration, and \(t\) is the time. On the other hand, Car 2's movement is constant speed, represented by the equation \(x_2(t) = 30.0 + vt\), where \(v\) is the constant speed, and \(t\) again denotes time.
These equations help set the stage for solving real-world problems. To find when two objects collide—like our two cars—setting the two positions equal allows us to solve for time and then use that time to determine the location.

• The two cars collide when their positions, \(x_1(t)\) and \(x_2(t)\), are equal. This gives us a common point to solve for, simplifying the problem to a straightforward algebraic equation.
• Each equation reflects the nature of motion specific to each car: constant speed for one and constant acceleration for the other, allowing both to be expressed in terms of their respective time-dependent positions.

Collision point calculation

The collision point calculation is essential to determine when and where two objects meet on their paths. By equating the expressions that describe each car's motion, we can calculate precisely where and when the collision occurs.
For this exercise, Car 1's position as a function of time is given by \(x_1(t) = \frac{1}{2} \times 2.0 \times t^2\), and Car 2's position is \(x_2(t) = 30.0 + 4.0t\). Setting these equal helps find the intersection point:
\(\frac{1}{2}(2.0)t^2 = 30.0 + 4.0t\)
This simplifies to:
\(t^2 = 30.0 + 4.0t\)
Solving further, we have:
\(t(t - 4.0) = 30.0\)
Because we seek a positive time value, solving this equation yields \(t = 5.0\) seconds. We then use this time in the first car's motion equation to find the collision along the track:
\(x_1(5.0) = \frac{1}{2}(2.0)(5.0)^2 = 25.0\, \mathrm{m}\).
This careful calculation shows where the first car meets the second. Understanding this allows us to visualize and quantify such dynamic scenarios, proving vital for various applications in physics and engineering.