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11.29: A.4.3- Section 4.3 Answers

  • Page ID
    121427
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    1. \(v=-\frac{384}{5}(1-e^{-5t/12});\quad -\frac{384}{5}\text{ft/s}\)

    2. \(k=12;\quad v=-16(1-e^{-2t})\)

    3. \(v=25(1-e^{-t});\quad 25\text{ft/s}\)

    4. \(v=20-27e^{-t/40}\)

    5. \(\approx 17.10\text{ ft}\)

    6. \(v=-\frac{40(13+3e^{-4t/5})}{13-3e^{-4t/5}};\quad -40\text{ft/s}\)

    7. \(v=-128(1-e^{-t/4})\)

    9. \(T=\frac{m}{k}\ln\left(1+\frac{v_{0}k}{mg} \right);\quad y_{m}=y_{0}=\frac{m}{k}\left[v_{0}-\frac{mg}{l}\ln\left(1+\frac{v_{0}k}{mg} \right) \right]\)

    10. \(v=-\frac{64(1-e^{-t})}{1+e^{-t}};\quad 64\text{ft/s}\)

    11. \(v=\alpha\frac{v_{0}(1+e^{-\beta t})-\alpha (1-e^{-\beta t})}{\alpha (1+e^{-\beta t})-v_{0}(1-e^{-\beta t})};\quad -\alpha\) where \(\alpha =\sqrt{\frac{mg}{k}}\) and \(\beta =2\sqrt{\frac{kg}{m}}\)

    12. \(T=\sqrt{\frac{m}{kg}}\tan ^{-1}\left(v_{0}\sqrt{\frac{k}{mg}} \right)\:v=-\sqrt{\frac{mg}{k}};\quad \frac{1-e^{-2\sqrt{\frac{ak}{m}}(t-T)}}{1+e^{-2\sqrt{\frac{ak}{m}}(t-T)}}\)

    13. \(s'=mg-\frac{as}{s+1};\quad a_{0}=mg\)

    14. (a) \(ms'=mg-f(s)\)

    15.

    1. \(v'=-9.8+v^{4}/81\)
    2. \(v_{T}\approx -5.308\text{m/s}\)

    16.

    1. \(v'=-32+8\sqrt{|v|};\quad v_{T}=-16\text{ ft/s}\)
    2. From Exercise 4.3.14
    3. \(v_{T}\) is the negative number such that \(-32 +8\sqrt{|v_{T}|}=0\); thus, \(v_{T}=-16\text{ ft/s}\)

    17. \(\approx 6.76\text{miles/s}\)

    18. \(\approx 1.47\text{miles/s}\)

    20. \(\alpha =\frac{gR^{2}}{(y_{m}+R)^{2}}\)


    This page titled 11.29: A.4.3- Section 4.3 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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