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If Pluto is 5.9x10^9 kilometers from Earth, how long will it take you, in years, to make the trip to Pluto and return home? (Assume a straight line, a constant velocity with no deceleration or acceleration, and a speed of 12.0 kilometers per second.)

*Hint: You will use the dimensional analysis conversions from earlier in the year to solve this!



Answer :

jacob193

Answer:

Approximately [tex]31[/tex] years, assuming that each year includes approximately [tex]365.25[/tex] days.

Explanation:

Make use of the following facts about the units:

  • [tex]1\; \text{hour} = 3600\; \text{second}[/tex].
  • [tex]1\; \text{day} = 24\; \text{hour}[/tex]
  • [tex]1\; \text{year} \approx 365.25\; \text{day}[/tex].

Divide total distance for a round-trip [tex]2 \times (5.9 \times 10^{9}\; {\rm km})[/tex] by speed to find the time required. As requested, leverage dimensional analysis to ensure that the unit of the result is in year. Make sure the intermediary units in the numerators match those in the denominators.

[tex]\begin{aligned}t =\; &\frac{s}{v} \\ =\; & \frac{2 \times (5.9 \times 10^{9}\; {\rm km})}{12 \; {\rm km\cdot s^{-1}}} \\ \approx\; &\frac{2 \times (5.9 \times 10^{9})}{12}\; {\text{s}} \\ &\times \frac{1\; \text{hour}}{3600\; \text{s}} \times \frac{1\; \text{day}}{24\; \text{hour}} \times \frac{1\; \text{year}}{365.25\; \text{day}} \\ =\; & \frac{2 \times (5.9 \times 10^{9})}{12 \times 3600 \times 24 \times 365.25}\; {\text{year}}\\ \approx\; & 31\; \text{year}\end{aligned}[/tex].

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