How Far Does a Body Travel in the Fourth Second If It Starts from Rest with a Uniform Acceleration of 2m/s2?
Introduction
This article delves into a common problem in physics: calculating the distance traveled by a body that starts from rest with a uniform acceleration. We will explore different methods and formulas to determine the precise distance traveled in the fourth second, using a simple acceleration of 2 m/s2. Whether you are an educator, a student, or simply someone with an interest in physics, this guide will provide a comprehensive understanding of the concepts involved.
Understanding the Problem
The problem statement provides a scenario where a body starts from rest (initial velocity, u 0 m/s) and accelerates uniformly at 2 m/s2. The goal is to find the distance traveled specifically in the fourth second. This problem can be solved using various methods, from straightforward calculations to the application of the SUVAT equations.
Solution Methods
Method 1: Straightforward Calculation
A simple and intuitive way to solve this problem is to calculate the distance traveled in each second and then find the difference between the distances traveled after the fourth and third seconds.
Distance Traveled After 4 Seconds: u 0 m/s a 2 m/s2 t 4 s(s ut 0.5at2)
(s 0 times 4 0.5 times 2 times 4^2 32) meters
(s ut 0.5at2)
(s 0 times 3 0.5 times 2 times 3^2 9) meters
Therefore, the distance traveled in the fourth second is:
(32 - 9 23) meters
Method 2: SUVAT Equations
The SUVAT equations are a set of five equations used to describe the motion of an object under constant acceleration. Here, we will use the equations to solve for the distance traveled in the fourth second.
Position Equations: (s ut 0.5at^2) (v u at) (v^2 u^2 2as) (s frac{v u}{2} times t) (v frac{s}{t} - frac{1}{2}at)Given:
u 0 m/s a 2 m/s2 t 4 sTo find the distance traveled in the fourth second, we first find the distance traveled after 4 seconds and then subtract the distance traveled after 3 seconds.
(s 0 times 4 0.5 times 2 times 4^2 32) meters (Total distance after 4 seconds)
(s 0 times 3 0.5 times 2 times 3^2 9) meters (Total distance after 3 seconds)
Distance traveled in the fourth second:
(32 - 9 23) meters
Using the Difference in Velocities
We can also calculate the distance traveled in the fourth second by using the concept that the velocity increases uniformly with time.
(v at) At the end of 4 seconds:
(v_4 2 times 4 8) m/s
At the end of 3 seconds:
(v_3 2 times 3 6) m/s
The distance traveled in the fourth second is the area under the velocity-time graph, which can be calculated as:
(s frac{v_4 v_3}{2} times (t_4 - t_3)) (s frac{8 6}{2} times (4 - 3) 7) meters
This method highlights the application of basic principles in physics.
Conclusion
The distance traveled in the fourth second by a body starting from rest with a uniform acceleration of 2 m/s2 is 7 meters. By using different methods and equations, we can arrive at the same conclusion. Understanding these concepts and formulas is crucial for solving more complex problems in physics and mechanics.
Keywords: distance traveled in second, uniform acceleration formula, suvat equations