Understanding the Relationship Between Distance and Velocity in Uniform Acceleration

Let's explore the fascinating world of physics through the lens of uniform acceleration. We'll delve into how we can use the equations of motion to solve intriguing problems involving cars, distances, and velocities. Our journey will begin by understanding the basic concepts and equations, followed by practical applications and real-world examples.

Introduction to Uniform Acceleration

Uniform acceleration is a critical concept in physics. When an object accelerates uniformly from rest, its velocity changes at a constant rate. This constant rate of change of velocity is represented as the acceleration, (a).

Uniform Acceleration Equations

The fundamental equations governing the motion of an object under uniform acceleration are quite powerful. They are:

( v^2 u^2 2as ) ( s ut frac{1}{2}at^2 ) ( v u at )

Here, (v) is the final velocity, (u) is the initial velocity, (a) is the acceleration, and (s) is the distance covered. These equations are the cornerstone of our analysis.

Problem Solving with Uniform Acceleration

Say a car accelerates uniformly from rest to reach a velocity (V) after covering a distance (D). We want to find the velocity of the car when it covers a distance of (2D).

First, we can use the equation: [ v^2 u^2 2as ]

Given that the car starts from rest, (u 0), and the final velocity after covering (D) is (V), the equation simplifies to:

[begin{align*}V^2 0 2aD V^2 2aD quad text{(1)}end{align*}]

From equation (1), we can solve for the acceleration (a): [begin{align*}a frac{V^2}{2D} quad text{(2)}end{align*}]

Next, we need to find the velocity when the car covers (2D). Using the same equation:

[begin{align*}v^2 u^2 2as v^2 0 2a(2D) v^2 4aD quad text{(3)}end{align*}]

Substituting the value of (a) from equation (2) into equation (3): [begin{align*}v^2 4 left(frac{V^2}{2D}right) D v^2 2V^2 quad text{(4)}end{align*}]

Finally, taking the square root of both sides, we get the desired velocity:

[begin{align*}v sqrt{2} Vend{align*}]

So, the velocity of the car when it covers a distance of (2D) is (sqrt{2} V).

Real-World Application

Consider a funny car that accelerates uniformly from the start line to a final velocity of 500 km/hr in 4.0 seconds. How far does it travel?

The relevant equation for this problem is:

[begin{align*}d frac{1}{2}at^2 quad text{(5)}end{align*}]

Given that (v at) and (v 500 , text{km/hr} frac{500 times 1000}{3600} , text{m/s} approx 138.89 , text{m/s}), we can find the acceleration (a):

[begin{align*}a frac{v}{t} a frac{138.89}{4} a approx 34.72 , text{m/s}^2end{align*}]

Substituting (a) and (t) into equation (5): [begin{align*}d frac{1}{2} (34.72) (4)^2 d frac{1}{2} (34.72) (16) d frac{1}{2} (555.52) d approx 277.76 , text{m}end{align*}]

Thus, the funny car travels approximately 277.76 meters.

Conclusion

In conclusion, we have demonstrated the application of uniform acceleration equations in solving problems related to distance and velocity. By understanding these equations and their derivations, we can tackle a wide range of real-world problems.

The velocity of a car when it covers a distance of (2D) after accelerating uniformly from rest to a velocity (V) over a distance (D) is given by (sqrt{2} V).