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derive-v2-u2-2as-graphically

🚀 Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In uniformly accelerated motion, the acceleration is constant, which leads to a linear relationship between the initial velocity, final velocity, acceleration, and displacement. The equation v² = u² + 2as describes this relationship, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. This equation can be derived graphically using a velocity-time graph, where the area under the graph represents displacement and the change in velocity is proportional to the acceleration.

Theory Explanation

Understanding the Velocity-Time Graph

In a velocity-time graph for uniformly accelerated motion, the velocity is plotted on the y-axis and time on the x-axis. The graph will be a straight line if the acceleration is constant. The slope of this line represents the acceleration (a). The area under the graph represents the displacement (s).

\[ s = \text{Area under the graph} \]
Deriving the Equation

The equation v = u + at describes the final velocity in terms of initial velocity, acceleration, and time. Rearranging gives us t = (v - u) / a. The area of the triangle formed in the graph can be calculated as (1/2) * base * height, which represents the change in velocity over time. Thus, the displacement can be expressed as s = ut + (1/2)at².

\[ s = ut + \frac{1}{2}at^2 \]
Substituting for Time

We can express time in terms of initial and final velocities. From v = u + at, we can derive t = (v - u) / a and substitute this into the displacement equation. After substituting and simplifying, we can arrive at the equation v² = u² + 2as.

\[ v^2 = u^2 + 2as \]

Key Points

  • 🎯 Kinematics deals with motion without considering forces.
  • 🎯 Uniformly accelerated motion has constant acceleration.
  • 🎯 The equation v² = u² + 2as relates initial and final velocities with acceleration and displacement.

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Examples:💡

An object starts from rest (u = 0 m/s) and accelerates at 2 m/s² over a distance of 50 m. Find the final velocity (v).

Solution:

Step 1: Using the equation v² = u² + 2as, substitute u = 0, a = 2 m/s², and s = 50 m.

\[ v^2 = 0^2 + 2 \cdot 2 \cdot 50 \]

Step 2: Calculate the right side: v² = 0 + 200 = 200.

\[ v^2 = 200 \]

Step 3: Take the square root to find v: v = √200 = 14.14 m/s.

\[ v = \sqrt{200} \]

A car moving at 20 m/s accelerates at 3 m/s². How far will it travel before reaching a speed of 35 m/s?

Solution:

Step 1: Use the equation v² = u² + 2as, where u = 20 m/s, v = 35 m/s, and a = 3 m/s².

\[ 35^2 = 20^2 + 2 \cdot 3 \cdot s \]

Step 2: Calculate: 1225 = 400 + 6s.

\[ 1225 = 400 + 6s \]

Step 3: Rearranging gives 6s = 825, so s = 137.5 m.

\[ s = \frac{825}{6} \]

Common Mistakes

  • Mistake: Confusing displacement with distance traveled; displacement is a vector quantity.

    Correction: Always remember that displacement has direction and is the shortest path between initial and final positions.

  • Mistake: Incorrectly applying the formula when acceleration is not constant.

    Correction: Ensure that the motion is uniformly accelerated before using the equation v² = u² + 2as.

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