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The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is?
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The average velocity of a body moving with uniform acceleration after ...
- Given Data
- Distance traveled (s) = 3.06 m
- Average velocity (v_avg) = 0.34 m/s
- Change in velocity (Δv) = 0.18 m/s
- Finding Acceleration
- Using the formula for average velocity: v_avg = (v_initial + v_final) / 2
- Given that the initial velocity (v_initial) is 0 m/s (since the body started from rest)
- We can rearrange the formula to find the final velocity (v_final) using the given average velocity
- v_final = 2 * v_avg - v_initial
- v_final = 2 * 0.34 - 0
- v_final = 0.68 m/s
- Calculating Acceleration
- Using the formula for acceleration: a = Δv / t
- Given that Δv = 0.18 m/s and t is the time taken to cover the distance of 3.06 m
- We can find t using the formula for uniform acceleration: s = (v_initial * t) + (1/2 * a * t^2)
- 3.06 = 0 * t + 0.5 * a * t^2
- t = sqrt((2 * 3.06) / a)
- Substituting the value of t in the acceleration formula, we get: a = 0.18 / sqrt((2 * 3.06) / a)
- Solving for Acceleration
- Solving the equation for acceleration, we get:
- a = 0.18 / sqrt(6.12 / a)
- a = 0.18 / sqrt(6.12) * sqrt(1/a)
- a = 0.18 / 2.47 * sqrt(1/a)
- a ≈ 0.0728 m/s^2
- Therefore, the uniform acceleration of the body is approximately 0.0728 m/s^2.
- Distance traveled (s) = 3.06 m
- Average velocity (v_avg) = 0.34 m/s
- Change in velocity (Δv) = 0.18 m/s
- Finding Acceleration
- Using the formula for average velocity: v_avg = (v_initial + v_final) / 2
- Given that the initial velocity (v_initial) is 0 m/s (since the body started from rest)
- We can rearrange the formula to find the final velocity (v_final) using the given average velocity
- v_final = 2 * v_avg - v_initial
- v_final = 2 * 0.34 - 0
- v_final = 0.68 m/s
- Calculating Acceleration
- Using the formula for acceleration: a = Δv / t
- Given that Δv = 0.18 m/s and t is the time taken to cover the distance of 3.06 m
- We can find t using the formula for uniform acceleration: s = (v_initial * t) + (1/2 * a * t^2)
- 3.06 = 0 * t + 0.5 * a * t^2
- t = sqrt((2 * 3.06) / a)
- Substituting the value of t in the acceleration formula, we get: a = 0.18 / sqrt((2 * 3.06) / a)
- Solving for Acceleration
- Solving the equation for acceleration, we get:
- a = 0.18 / sqrt(6.12 / a)
- a = 0.18 / sqrt(6.12) * sqrt(1/a)
- a = 0.18 / 2.47 * sqrt(1/a)
- a ≈ 0.0728 m/s^2
- Therefore, the uniform acceleration of the body is approximately 0.0728 m/s^2.
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The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is?
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The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is?.
The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 M is 0.34 M per second if the change in velocity of the body is 0.18 M per second during this time is uniform acceleration is?.
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