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derive-v-u-at-graphically

🚀 Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. One important aspect of kinematics is uniformly accelerated motion, where an object moves with a constant acceleration. This means that the object's velocity changes at a constant rate over time. In this context, we can derive the relationship between initial velocity (u), final velocity (v), acceleration (a), and time (t) graphically using a velocity-time graph. The area under the velocity-time graph represents the displacement of the object, and the slope of the graph represents the acceleration.

Theory Explanation

Understanding Velocity-Time Graphs

In a velocity-time graph, the x-axis represents time (t), and the y-axis represents velocity (v). For uniformly accelerated motion, the graph is a straight line, indicating that velocity changes at a constant rate. The slope of this line represents acceleration (a). If the line is sloping upwards, the object is accelerating; if it is sloping downwards, the object is decelerating.

Deriving the Equation v = u + at

From the velocity-time graph, we can see that the change in velocity (v - u) over time (t) is equal to the acceleration (a). Mathematically, this can be expressed as v - u = at, which can be rearranged to give the equation v = u + at. This equation shows how the final velocity (v) depends on the initial velocity (u), the acceleration (a), and the time (t) over which the acceleration occurs.

\[ v = u + at \]
Understanding the Area Under the Graph

The area under the velocity-time graph represents the displacement (s) of the object. For a uniformly accelerated motion, the area can be calculated as the area of a trapezoid or a rectangle, depending on the situation. The formula for displacement can also be derived from the graph, leading to the equation s = ut + (1/2)at².

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

Key Points

  • 🎯 Kinematics deals with motion without considering forces.
  • 🎯 Uniformly accelerated motion means constant acceleration.
  • 🎯 The slope of the velocity-time graph represents acceleration.

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Examples:💡

An object starts from rest and accelerates at a rate of 2 m/s² for 5 seconds. Find the final velocity and displacement.

Solution:

Step 1: Given initial velocity u = 0 m/s, acceleration a = 2 m/s², and time t = 5 s. Use the equation v = u + at to find the final velocity.

\[ v = 0 + (2)(5) = 10 m/s \]

Step 2: Now, use the equation s = ut + (1/2)at² to find the displacement.

\[ s = (0)(5) + \frac{1}{2}(2)(5^2) = 0 + 25 = 25 m \]

A car moving at 20 m/s accelerates at 3 m/s² for 4 seconds. Calculate the final velocity and the distance traveled during this time.

Solution:

Step 1: Given initial velocity u = 20 m/s, acceleration a = 3 m/s², and time t = 4 s. Use v = u + at to find the final velocity.

\[ v = 20 + (3)(4) = 20 + 12 = 32 m/s \]

Step 2: Now, use s = ut + (1/2)at² to find the displacement.

\[ s = (20)(4) + \frac{1}{2}(3)(4^2) = 80 + 24 = 104 m \]

Common Mistakes

  • Mistake: Confusing acceleration with velocity; acceleration is the rate of change of velocity, not the velocity itself.

    Correction: Always remember that acceleration indicates how quickly the velocity is changing.

  • Mistake: Forgetting to convert units when calculating; for example, mixing m/s with km/h.

    Correction: Always ensure that units are consistent before performing calculations.

  • Mistake: Neglecting the initial velocity when using the formulas; students sometimes assume initial velocity is zero.

    Correction: Read the problem carefully to determine the correct initial velocity.

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