Question

A car starts from rest and accelerates at \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). How far does it travel in \(2.00 \mathrm{~s} ?\)

Short answer

Answer: The car travels a distance of 20.0 m in 2.00 s.

Step-by-step solution

Write down given values and equation

We are given: - Initial position (\(x_0\)): 0 - Initial velocity (\(v_0\)): 0 - Acceleration (\(a\)): \(10.0 \mathrm{~m} / \mathrm{s}^{2}\) - Time (\(t\)): \(2.00 \mathrm{~s}\) The equation for the distance traveled (\(x\)) is: \(x = x_0 + v_0t+\frac{1}{2}at^2\)

Substitute the given values into the equation

Plug in the given values for the initial position, initial velocity, acceleration, and time into the equation: \(x = 0 + 0\times2.00+\frac{1}{2}(10.0)(2.00)^2\)

Calculate the distance traveled

Now simplify and calculate the distance traveled: \(x = \frac{1}{2}(10.0)(4.00)\) \(x = 5.0\times4.00\) \(x = 20.0 \mathrm{~m}\) The car travels a distance of \(20.0 \mathrm{~m}\) in \(2.00 \mathrm{~s}\).

Key concepts

Acceleration

When we talk about acceleration, we are referring to the rate at which an object changes its velocity. It is a vector quantity, meaning it has both magnitude and direction. Acceleration can be due to a change in speed or direction, or both. In the case of our exercise, the car's acceleration is constant at 10.0 meters per second squared (m/s²). This tells us that every second, the car's velocity increases by 10.0 m/s. Understanding acceleration is crucial because it's the key factor that affects how the velocity of an object changes over time.

In kinematics, acceleration is often one of the known parameters when solving problems, alongside initial velocity, final velocity, time taken, and distance covered. When an object starts from rest, such as the car in our exercise, its initial velocity is zero, which simplifies the calculation of distance traveled.

Distance Traveled

The distance traveled by an object is the total length of its path during a specified time interval. For uniformly accelerated motion, where acceleration is constant, we can use specific equations to find this distance. In our car scenario, we want to know how far the car travels when it accelerates from rest for 2.00 seconds. Remember, when working with kinematic problems, it's essential to keep track of units to ensure accuracy.

It's important to distinguish distance traveled from displacement, which is the change in position. In a straight-line motion with constant acceleration starting from rest, these two values are the same, but they can differ if the path isn't straight or if the object isn't starting from rest.

Equations of Motion

The equations of motion are the bread and butter of kinematics. These equations allow us to relate displacement, initial and final velocities, acceleration, and time. They are essential tools when we need to predict future motion or back-calculate aspects of past motion given certain data points. In our exercise, we use the equation \( x = x_0 + v_0t + \frac{1}{2}at^2 \) to find the distance traveled. This is one of the several standard equations used in kinematics, often referred to as the 'suvat' equations (s for displacement, u for initial velocity, v for final velocity, a for acceleration, and t for time).

These equations are based on the assumption of uniform acceleration and only apply to such scenarios. Students should practice these equations frequently to become proficient in selecting the right one for each problem they face.

Uniformly Accelerated Motion

Uniformly accelerated motion occurs when an object’s acceleration is constant in both magnitude and direction. In the context of our exercise, this means the car's speed increases by the same amount every second. A key point to remember is that in uniformly accelerated motion, the final velocity is not needed to calculate the distance traveled, because the acceleration doesn't change. This makes problems of this type more straightforward.

When dealing with uniformly accelerated motion, visualization can be quite helpful; for example, imagine a graph where you plot velocity against time. With uniform acceleration, the graph would be a straight line, where the slope represents the acceleration. Knowing this can also help students understand why the formula for the distance includes the term \( \frac{1}{2}at^2 \) - it's the area under the acceleration-time graph, representing the total distance traveled.