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prove- v=u+at, s=ut+1/2at2, v2-u2=2as
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prove- v=u+at, s=ut+1/2at2, v2-u2=2as
Introduction:
In this explanation, we will prove the equations v = u + at, s = ut + 1/2at^2, and v^2 - u^2 = 2as. These equations are fundamental kinematic equations used to describe the motion of an object in one dimension. We will break down each equation and provide a detailed explanation for each.
Proofs:
1. v = u + at:
This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and time (t). Let's prove it step by step:
- Step 1: Consider an object with an initial velocity u and acceleration a. After time t, the object reaches a final velocity v.
- Step 2: During this time interval, the object experiences a change in velocity (Δv), which is given by the equation Δv = v - u.
- Step 3: In the given time t, the object undergoes a constant acceleration a. Therefore, the average acceleration (a_avg) can be defined as a_avg = Δv / t.
- Step 4: Rearranging the equation from step 3, we find Δv = a_avg * t.
- Step 5: Since the acceleration is constant, a_avg is equal to a. Therefore, Δv = a * t.
- Step 6: Substituting the value of Δv from step 2, we have v - u = a * t.
- Step 7: Finally, rearranging this equation gives us v = u + at, which is the desired result.
2. s = ut + 1/2at^2:
This equation represents the displacement (s) of an object in terms of its initial velocity (u), time (t), and acceleration (a). Let's prove it step by step:
- Step 1: Consider an object with an initial velocity u, acceleration a, and time t. We want to find the displacement (s) of the object during this time interval.
- Step 2: The displacement (s) can be defined as the area under the velocity-time graph.
- Step 3: The velocity-time graph for an object with constant acceleration a is a straight line.
- Step 4: The area under a velocity-time graph is given by the formula A = base * height. In this case, the base is t and the height is (u + v)/2.
- Step 5: Substituting the values, we have s = t * (u + v)/2.
- Step 6: Using the equation v = u + at (proved in step 1), we can replace v with (u + at).
- Step 7: Simplifying the equation, we get s = ut + 1/2at^2, which is the desired result.
3. v^2 - u^2 = 2as:
This equation relates the final velocity
In this explanation, we will prove the equations v = u + at, s = ut + 1/2at^2, and v^2 - u^2 = 2as. These equations are fundamental kinematic equations used to describe the motion of an object in one dimension. We will break down each equation and provide a detailed explanation for each.
Proofs:
1. v = u + at:
This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and time (t). Let's prove it step by step:
- Step 1: Consider an object with an initial velocity u and acceleration a. After time t, the object reaches a final velocity v.
- Step 2: During this time interval, the object experiences a change in velocity (Δv), which is given by the equation Δv = v - u.
- Step 3: In the given time t, the object undergoes a constant acceleration a. Therefore, the average acceleration (a_avg) can be defined as a_avg = Δv / t.
- Step 4: Rearranging the equation from step 3, we find Δv = a_avg * t.
- Step 5: Since the acceleration is constant, a_avg is equal to a. Therefore, Δv = a * t.
- Step 6: Substituting the value of Δv from step 2, we have v - u = a * t.
- Step 7: Finally, rearranging this equation gives us v = u + at, which is the desired result.
2. s = ut + 1/2at^2:
This equation represents the displacement (s) of an object in terms of its initial velocity (u), time (t), and acceleration (a). Let's prove it step by step:
- Step 1: Consider an object with an initial velocity u, acceleration a, and time t. We want to find the displacement (s) of the object during this time interval.
- Step 2: The displacement (s) can be defined as the area under the velocity-time graph.
- Step 3: The velocity-time graph for an object with constant acceleration a is a straight line.
- Step 4: The area under a velocity-time graph is given by the formula A = base * height. In this case, the base is t and the height is (u + v)/2.
- Step 5: Substituting the values, we have s = t * (u + v)/2.
- Step 6: Using the equation v = u + at (proved in step 1), we can replace v with (u + at).
- Step 7: Simplifying the equation, we get s = ut + 1/2at^2, which is the desired result.
3. v^2 - u^2 = 2as:
This equation relates the final velocity
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prove- v=u+at, s=ut+1/2at2, v2-u2=2as
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prove- v=u+at, s=ut+1/2at2, v2-u2=2as for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about prove- v=u+at, s=ut+1/2at2, v2-u2=2as covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for prove- v=u+at, s=ut+1/2at2, v2-u2=2as.
prove- v=u+at, s=ut+1/2at2, v2-u2=2as for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about prove- v=u+at, s=ut+1/2at2, v2-u2=2as covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for prove- v=u+at, s=ut+1/2at2, v2-u2=2as.
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