How to Calculate the Time for a Car to Travel 50 Meters from Rest
When analyzing the dynamics of motion, one fundamental question arises: how long does it take for a car to travel 50 meters if it starts from rest with a constant acceleration of 5 m/s2? This topic is not only interesting but also crucial for understanding basic kinematics and physics. In this article, we will walk through the process of solving this problem step-by-step using the kinematic equation and provide a clear understanding of each step involved.
Understanding the Problem
The car starts from rest, meaning its initial velocity (v_0) is 0 m/s. It accelerates at a constant rate of 5 m/s2. We need to determine the time (t) it takes for the car to travel a distance of 50 meters.
Using the Kinematic Equation
The kinematic equation for distance traveled under constant acceleration is given by:
[d v_0 t frac{1}{2} a t^2]
Given that (v_0 0) m/s, the equation simplifies to:
[d frac{1}{2} a t^2]
Simplifying the Equation
Substituting the known values into the equation, we get:
[50 frac{1}{2} times 5 times t^2]
This further simplifies to:
[50 frac{5}{2} t^2]
Solving for Time
To isolate (t^2), we can multiply both sides of the equation by 2:
[100 5 t^2]
Next, we divide both sides by 5:
[t^2 20]
Finally, taking the square root of both sides, we find:
[t sqrt{20} approx 4.47 text{ seconds}]
A Visual Explanation
The distance traveled can also be viewed as the area under a velocity-time graph. Since the acceleration is constant, the graph is a straight line. The area under this graph forms a triangle, where the height is the acceleration multiplied by time, and the base is the time duration.
The area of the triangle can be calculated using the formula:
[d frac{1}{2} a t^2]
Plugging in the values, we get:
[50 frac{1}{2} times 5 times t^2]
This simplifies to the same equation as before, confirming our previous calculations.
Additional Considerations and Clarifications
It's important to note that the unit of acceleration is m/s2, which clearly indicates the change in velocity per unit of time, not m/s. Acceleration is a vector quantity, meaning it has both magnitude and direction.
While the simplified problem presented here assumes constant acceleration in a single direction, in more complex scenarios involving different directions or varying acceleration, the problem becomes more intricate and may require vector analysis.
For further reading or detailed calculations on similar problems, you can refer to online resources or Wikipedia articles on kinematics. These resources often provide a deeper understanding of the underlying principles and offer more detailed explanations and examples.
Understanding the basics of kinematics is not only important for physics students but also for engineering and other practical applications where motion analysis is essential.
For additional resources or further inquiries, consider exploring websites like PhysicsClassroom or Khan Academy.