Understanding Uniform Acceleration: Calculating Acceleration and Distance Traveled
Understanding the principles of uniform acceleration is essential in various fields, from physics to engineering. In this article, we will explore the dynamics of an object starting from rest and accelerating at a constant rate, specifically focusing on calculating the acceleration and distance traveled over a given period. We will use kinematic equations to solve for unknowns and interpret the results.
Uniform Acceleration Basics
Uniform acceleration refers to the situation where an object moves with a constant acceleration. This means that the velocity of the object changes by a constant amount every second. The basic kinematic equations help us solve for various parameters, such as distance, time, and acceleration. These equations are:
( s ut frac{1}{2}at^2 ) ( v^2 u^2 2as ) ( v u at ) ( s frac{u v}{2} t )Problem Setup
Consider an object starting from rest, experiencing uniform acceleration, and traveling a distance of 100 meters in 5 seconds. We need to find the value of the acceleration and the velocity as a function of time.
Solving for Acceleration
We start with the kinematic equation for distance covered under uniformly accelerated motion:
[ s ut frac{1}{2}at^2 ]Given that the initial velocity ( u 0 ) and the distance ( s 100 ) meters in ( t 5 ) seconds, we substitute these values into the equation:
[ 100 0 cdot 5 frac{1}{2} a cdot 5^2 ]Performing the calculations, we get:
[ 100 frac{25}{2} a ]Therefore,
[ a frac{100 times 2}{25} 8 ms^{-2} ]So, the acceleration is ( 8 ms^{-2} ).
Distance Traveled in Any Second
Next, we calculate the distance traveled in the 5th second. To do this, we need to find the distance traveled in the first 5 seconds and the first 4 seconds. The difference between these two will give us the distance traveled in the 5th second.
The distance traveled in 5 seconds is given by:
[ s_5 frac{1}{2} a cdot 5^2 frac{1}{2} cdot 8 cdot 25 100 text{meters} ]The distance traveled in 4 seconds is:
[ s_4 frac{1}{2} a cdot 4^2 frac{1}{2} cdot 8 cdot 16 64 text{meters} ]Thus, the distance traveled in the 5th second is:
[ s_5 - s_4 100 - 64 36 text{meters} ]Note that the provided value in the problem statement, 3 meters, is an error in the problem setup. Using the correct kinematic equations, we find that the distance traveled in the 5th second is 36 meters.
Velocity as a Function of Time
Finally, we find the velocity as a function of time. Using the kinematic equation ( v u at ), where ( u 0 ) and ( a 8 ms^{-2} ):
[ v 0 8t 8t text{m/s} ]This means the velocity increases linearly with time, starting from 0 at ( t 0 ) and reaching 8 meters per second after 1 second, 16 meters per second after 2 seconds, and so on.
Conclusion
In this article, we utilized the principles of uniform acceleration and kinematic equations to solve for the acceleration and distance traveled over a given period. Understanding these concepts is crucial for various applications in physics and engineering. By mastering these fundamental principles, we can accurately describe and predict the motion of objects in a wide range of scenarios.