Understanding Acceleration: Calculating Velocity for a Car Starting from Rest

When a car starts from rest and accelerates at a constant rate, the changes in its velocity can be understood through basic physics principles. This article explains the process of calculating the velocity of the car after a certain period of time, using the example of a car that accelerates at 5.0 m/s2 for 3.0 seconds.

Introduction to Acceleration

Acceleration, fundamentally, is defined as the rate of change of velocity with respect to time. It tells us how quickly the velocity of an object is changing. In this case, the car is accelerating at 5.0 meters per second squared (m/s2), which means its velocity increases by 5.0 m/s every second.

Calculating Velocity

To calculate the velocity of the car after 3.0 seconds, we can use the basic formula:

v aT

v final velocity a constant acceleration (5.0 m/s2) T time (3.0 seconds)

Substituting the values, we get:

v 5.0 m/s2 x 3.0 s 15.0 m/s

Therefore, the velocity of the car after 3.0 seconds is 15.0 m/s.

Alternative Method Using Equations

Another method to solve this problem is by using the kinematic equations. We start with the following equation:

s ut 0.5at2

s distance u initial velocity (0 m/s since the car starts from rest) a acceleration (5.0 m/s2) t time (3.0 seconds)

Substituting the values, we get:

s 0.5 x 5.0 m/s2 x (3.0 s)2 22.5 meters

This equation provides the distance traveled by the car. However, to find the final velocity, we use the formula:

vfinal u at

vfinal final velocity u initial velocity (0 m/s) a acceleration (5.0 m/s2) t time (3.0 seconds)

Substituting the values, we get:

vfinal 0 m/s 5.0 m/s2 x 3.0 s 15.0 m/s

Interpreting the Results

The results clearly show that the car's velocity after 3.0 seconds is 15.0 m/s. It is crucial to understand that this process can be simplified using the fundamental principles of acceleration, making it easier to comprehend and apply in real-world scenarios.

Additional Insights

If the question were framed differently, such as asking about the distance traveled by the car, then the displacement would be the area under the velocity-time graph. For the given scenario, the displacement can be calculated as the area of a triangle:

s vavg x t

vavg vi vf/2

vi 0 m/s (initial velocity)

vf vi at

vf 0 m/s 5.0 m/s2 x 3.0 s 15.0 m/s

vavg (0 m/s 15.0 m/s) / 2 7.5 m/s

s 7.5 m/s x 3.0 s 22.5 meters

Conclusion

Understanding and calculating velocity and acceleration is a fundamental aspect of physics. The methods discussed here provide a clear and straightforward approach to solving such problems. Whether you use the formula or the kinematic equations, the key is to break down the problem and understand the underlying principles.

By mastering these concepts, students can confidently answer questions about motion and acceleration. Whether it's for homework, a test, or a real-world application, a solid grasp of physics principles will serve as a valuable tool.