Calculating Distance Covered by a Body with Initial Velocity and Acceleration

Understanding the motion of a body under constant acceleration is crucial in physics and engineering. One of the simplest scenarios involves a body starting from rest with an initial velocity, accelerating at a constant rate, and then either continuing to move or decelerating to a stop. This article will guide you through the process of calculating the distance covered by such a body over a given period of time. We will use the kinematic equations and explore common acceleration formulas.

Kinematic Equations and Their Application

The kinematic equations are fundamental in understanding motion with constant acceleration. Two of the key equations used are:

Distance: s ut 0.5at^2 Velocity: v u at

These equations can be used together to solve a variety of motion problems. Let’s apply these to a real scenario to better understand their use.

Example Scenario

Consider a body starting from rest with an initial velocity of 10 m/s, and acceleration of 5 m/s^2. We want to find the distance covered by the body in 5 seconds.

Step-by-Step Solution

Given:

Initial velocity (v_i): 10 m/s Acceleration (a): 5 m/s^2 Time (t): 5 s

To find the distance (s) covered, we will use the distance formula:

s ut 0.5at^2

Substituting the given values:

u (initial velocity): 10 m/s a (acceleration): 5 m/s^2 t (time): 5 s

s 10 times 5 0.5 times 5 times 5^2

Calculating each term:

ut: 10 times 5 50 m 0.5at^2: 0.5 times 5 times 25 62.5 m

Adding both parts together:

s 50 62.5 112.5 m

Thus, the distance covered in 5 seconds is 112.5 meters.

Understanding Acceleration Formulas

There are a few essential formulas related to acceleration that can be used to solve motion problems:

Velocity: v u at Distance (Constant Initial Velocity): s ut 0.5at^2 Final Velocity with Deceleration: v^2 u^2 2as

These formulas are interrelated and can be used in conjunction to solve more complex motion problems. For example, if a body starts from rest and accelerates for a certain period, then decelerates, you can use these formulas to find the distance covered in each phase.

Example with Deceleration

Let's consider a more complex scenario where the body initially accelerates for 10 seconds, then decelerates to a stop.

First Phase: Acceleration

Given:

u (initial velocity): 0 m/s (assuming starting from rest) a (acceleration): 10 m/s^2 t (time): 10 s

Using the distance formula:

s ut 0.5at^2 0 0.5 times 10 times 100 500 m

Velocity after 10 seconds:

v u at 0 10 times 10 100 m/s

Second Phase: Deceleration

Now, the body decelerates to a stop over the next 10 seconds. Given:

u (initial velocity): 100 m/s a (deceleration): -5 m/s^2 v (final velocity): 0 m/s (body comes to a stop)

Using the distance formula:

v^2 u^2 2as implies 0 100^2 2 times (-5) times s implies 10000 10s implies s 1000 m

Thus, the distance covered during deceleration is 1000 meters.

The total distance covered is:

500 m 1000 m 1500 m

In conclusion, the total distance covered in both phases is 1500 meters.

Understanding these fundamental concepts and formulas helps in solving complex motion problems in physics and engineering. Whether you need to determine distance covered during acceleration or deceleration, the kinematic equations and acceleration formulas provide a powerful toolkit.