Understanding Uniform Acceleration in a Car
The problem of determining the acceleration of a car that starts from rest and covers a distance of 100 meters in one second with uniform acceleration can be solved using kinematic equations. This article will explore the various methods to solve for the acceleration, including the use of the kinematic formula S frac{1}{2} at^2, as well as analyze the v-t diagram approach and Galileo's Rule of Odd Intervals.
Kinematic Equations and the S frac{1}{2} at^2 Method
Given the car starts from rest, the initial velocity u 0 , text{m/s}, the time t 6 , text{s}, and the distance S 100 , text{m}, we can use the kinematic equation S ut frac{1}{2} at^2 to solve for the acceleration a.
Substituting the known values:
100 , text{m} 0 cdot 6 frac{1}{2} a cdot 6^2Simplifying the equation:
100 , text{m} frac{1}{2} a cdot 36 , text{s}^2frac{100}{36} , text{m/s}^2 a Rightarrow a frac{100}{18} , text{m/s}^2
The acceleration of the car is approximately 5.556 m/s2 or frac{50}{9} , text{m/s}^2.
Alternative Method: The v-t Diagram and Distance Calculation
Another approach to determine the acceleration is by analyzing the object's motion using a velocity-time (v-t) diagram. For a motion starting from rest with constant acceleration, the v-t diagram forms a triangle starting at 0 m/s and rising to the final velocity. The total displacement can be calculated using the area of the v-t diagram.
Given the total distance traveled S 100 , text{m} and time t 6 , text{s}, this can be broken down into two parts: the first second and the subsequent seconds. The first second covers a known distance, and the remaining 95 meters are covered in the next 5 seconds.
The area under the v-t diagram for the second to the sixth second is a trapezoid. The area of the trapezoid can be calculated using the equation:
S frac{1}{2} (v_3 u) cdot 3 , text{s}Since u 0 , text{m/s} and the final velocity at the end of the sixth second is 10 m/s, we have:
100 - distance covered in the first second frac{1}{2} (10 0) cdot 3Therefore, the distance covered in the first second can be calculated by subtracting 15 meters from 100 meters:
distance covered in the first second 100 , text{m} - 15 , text{m} 85 , text{m}The acceleration can be found by using the relationship between distance, initial velocity, and time:
S ut frac{1}{2} at^2 Rightarrow 85 , text{m} 0 frac{1}{2} a cdot 1 , text{s}^2 Rightarrow a 170 , text{m/s}^2 / 1 , text{s}^2 170 , text{m/s}^2This approach involves multiple steps and assumptions but provides a clear visualization of the car's motion.
Galileo's Rule of Odd Intervals
A simpler method to solve the problem using Galileo's Rule of Odd Intervals is applicable when the motion is uniform acceleration. According to this rule, the distance traveled during consecutive intervals of equal times increases in the ratio 1:3:5:7:9. In this case, the motion is broken down into intervals of one second, and the fourth second covers a distance of 21 meters.
Using Galileo's Rule, the fourth second will cover the fourth term in the series, which is 7 meters. Therefore, the acceleration can be found as:
a frac{7}{2 cdot 4 cdot 4} frac{7}{32} , text{m/s}^2 cdot 2 approx 6 , text{m/s}^2This method provides an approximate solution and is useful for quick calculations.
Conclusion
Understanding the kinematic equations and various methods to solve for acceleration, such as the S frac{1}{2} at^2 formula, v-t diagram, and Galileo's Rule of Odd Intervals, allows for a comprehensive analysis of uniform acceleration problems. The acceleration of the car in this particular problem is approximately 5.556 m/s2 or frac{50}{9} , text{m/s}^2, providing a consistent and accurate result.