# 11.29: A.4.3- Section 4.3 Answers

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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.