Understanding the Relationship Between Distance Traveled in Successive Intervals of Constant Acceleration
When a body is in motion with a constant acceleration, the distances traveled in successive intervals can be analyzed using the principles of motion and the equations of motion. These equations allow us to understand how the distance covered in the first interval compares to the distance covered in a subsequent interval. This article will explore the mathematical relationship between distances traveled, S1 and S2, under conditions of constant acceleration.
Introduction to Constant Acceleration and Kinematic Equations
When a particle experiences a constant acceleration (a), its velocity (v) and displacement (x) can be expressed through the equations of motion. These equations are derived from the fundamental principles of kinematics, which relate the displacement, velocity, acceleration, and time of a moving object. The SUVAT equations, named after the symbols used (s: displacement, u: initial velocity, v: final velocity, a: acceleration, t: time), are particularly useful in this context.
Equation of Motion for Distance Traveled in First and Second Intervals
Consider a body moving with a constant acceleration of (10 , text{m/s}^2). The distances traveled in the first (S1) and second seconds (S2) can be calculated using the equations of motion. The displacement (s) of a particle in the first (t) seconds is given by the formula:
[ s frac{1}{2} a t^2 ]
To find the distance traveled in the first second:
[ S1 frac{1}{2} times 10 times 1^2 5 , text{meters} ]
At the end of the second second, the velocity of the body is (v u at 0 10 times 2 20 , text{m/s}), and the distance traveled in the second second is:
[ S2 frac{1}{2} times 10 times 2^2 - frac{1}{2} times 10 times 1^2 20 - 5 15 , text{meters} ]
Thus, we see that:
[ S2 3 times S1 ]
This relationship arises because the distance traveled in the second second is composed of the displacement in the first second plus the additional distance covered in the second second.
Analysis for Different Time Intervals
For a time period of 10 seconds, the distance covered, S1, can be calculated as:
[ S1 frac{1}{2} times a times 10^2 50 times a ]
For the next 10 seconds (S2), the distance covered is:
[ S2 frac{1}{2} times a times 20^2 - frac{1}{2} times a times 10^2 200a - 50a 150a ]
Therefore:
[ frac{S2}{S1} frac{150a}{50a} 3 ]
Generalized Relationship and Application
In a more generalized form, the relationship can be expressed for any given time period. If the acceleration is (a), the distance traveled in the first (n) seconds (S1) and the next (n) seconds (S2) is given by:
[ S1 frac{1}{2} times a times n^2 ]
[ S2 frac{1}{2} times a times (2n)^2 - frac{1}{2} times a times n^2 a times 2n^2 - frac{1}{2} a times n^2 frac{3}{2} a n^2 ]
Thus, the ratio S2/S1 is:
[ frac{S2}{S1} frac{3}{2} ]
This explains why the distance traveled in the second interval is always 3 times the distance traveled in the first interval, as long as the intervals are of equal duration.
Understanding these relationships is essential for analyzing the motion of particles under constant acceleration, which is a fundamental concept in physics and engineering. These principles can be applied to a wide range of scenarios, from simple physics problems to more complex engineering applications.