An object in free fall is at heights y1, y2, and y3 at times t1, t2, and t3 respectively.
Part (a) Which of the following is the correct algebraic expression for the average velocity v12 of the object as it moves from y1 to y2?
ANSWER: (y2-y1)/(t2-t1)
Part (b) Which of the following is the correct expression for the midpoint of the time interval t12 at which the instantaneous velocity is exactly equal to this average velocity?
ANSWER: (t1+t2)/2
Part (c) Based on your answers to parts (a) and (b), you can now write similar expressions for the average velocity v23, and the midpoint t23 of the time interval from t2 to t3. Use your answers from parts (a) and (b), and the above information to write the algebraic expression for the average acceleration for these two time intervals. (Use the following as necessary: t12, t23, v12, v23.)
aaverage = _____________

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moya1316 moya1316 · Answer 1 · 2020-09-18T02:44:54+02:00

Answer:

a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

Explanation:

This is an exercise of average speed, which is defined with the variation of the distance in the unit of time

v = (y₃ - y₂) / (t₃-t₂)

the midpoint of a magnitude is the sum of the magnitude between 2

t_mid = (t₂ + t₃) / 2

the same reasoning is used for the mean acceleration

a = (v_f - v₀) / (t_f - t₀)

in our case

a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)