Question

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

Short answer

Answer: The car travels a distance of 20.0 meters in 2.00 seconds.

Step-by-step solution

List given information

The problem provides us with the following information: - The car starts from rest, meaning its initial velocity is 0: \(v_0 = 0 \mathrm{~m}/\mathrm{s}\) - The car accelerates at a constant rate: \(a = 10.0 \mathrm{~m}/\mathrm{s}^{2}\) - We need to find the distance traveled in a given time: \(t = 2.00 \mathrm{~s}\)

Use the kinematic equation to determine the distance traveled

Since we have the initial velocity, acceleration, and time, we can use the following kinematic equation to find the distance traveled: $$ d = v_0t + \frac{1}{2}at^2 $$ This equation states that the distance traveled (d) is equal to the product of the initial velocity (v_0) and time (t) plus one-half the product of the acceleration (a) and the square of the time (t^2).

Substitute the given values and solve for the distance

Substitute the given values into the equation: $$ d = (0 \mathrm{~m}/\mathrm{s})(2.00 \mathrm{~s}) + \frac{1}{2}(10.0 \mathrm{~m}/\mathrm{s}^{2})(2.00 \mathrm{~s})^2 $$ Now, simplify the equation and solve for the distance: $$ d = 0 + (10.0 \mathrm{~m}/\mathrm{s}^{2})(4.00 \mathrm{~s}^2)/2 $$ $$ d = (20.0 \mathrm{~m}/\mathrm{s}^{2})(2.00 \mathrm{~s}^2) $$ $$ d = 20.0 \mathrm{~m} $$

State the final answer

The car travels a distance of \(20.0\) meters in \(2.00\) seconds.

Key concepts

Understanding Acceleration

Acceleration is a key concept in kinematics and describes how the velocity of an object changes over time. In simple terms, acceleration means speeding up or slowing down. It is a vector quantity, which means it has both magnitude and direction. For example, if a car accelerates at \(10.0 \, \mathrm{m/s^2}\), this shows the car's speed increases by \(10.0 \, \mathrm{m/s}\) every second.

Here's a quick snapshot of acceleration:

• Measured in meters per second squared (\(\mathrm{m/s^2}\)).
• Positive acceleration means increasing velocity.
• Negative acceleration (or deceleration) means decreasing velocity.

In the exercise, the acceleration of the car is given as \(10.0 \, \mathrm{m/s^2}\), indicating how quickly it gains speed from rest. It is important to remember that this value remains constant over the time period used in kinematics calculations.

Calculating Distance Traveled

The distance traveled by an object can be calculated using information about its initial velocity, acceleration, and the time it has been in motion. In our scenario, the car starts from rest, meaning the initial velocity is zero. The goal is to determine how far it travels after accelerating for a specific duration of time.

To find the distance, follow these steps:

• Use the kinematic equation which helps relate initial velocity, acceleration, time, and distance.
• Substitute known values: the initial velocity \((v_0)\), acceleration \((a)\), and time \((t)\).
• Perform the calculations step-by-step as shown in the solution to determine the total distance.

Understanding how to apply these steps is crucial for solving a range of physics problems involving moving objects.

The Kinematic Equation Explained

Kinematics often involves equations that relate different physical quantities. One of the foundational kinematic equations involves the displacement of an object when initial velocity, acceleration, and time are known. The equation used here is:\[ d = v_0t + \frac{1}{2}at^2\]This equation provides a way to calculate the distance \((d)\) based on:

• Initial velocity \((v_0)\): The starting speed of the object.
• Acceleration \((a)\): The change in velocity per time unit.
• Time \((t)\): How long the object has been accelerating.

In this case, the initial velocity \((v_0)\) is zero, simplifying our calculations since the first term, \(v_0t\), becomes zero. Thus, the equation simplifies to calculating \(\frac{1}{2}at^2\). It's important to remember that the time \((t)\) must be squared, which gives the formula its distinctive form and requires careful attention when solving problems."}]}}]}}]}]}