Assume that yo-yo drops 1.0 m vertically from its owner's hand at a uniform speed of 0.80 m/s along its string. At the end of the string, it changes direction in 0.10 s and rises toward the owner's hand at a uniform speed of 0.60 m/s. Compute the average acceleration of the yo-yo at the bottom end of the string as it changes direction.
Answer
acceleration is that
"a=\\frac{v-v'}{t-t'}\\\\=\\frac{0.8+0.6}{0.1}\\\\=14m\/s^2"
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