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While I was studying about finite differences I noticed a question in difference equations.Does anyone knows how to solve this using difference equations?

WORD PROBLEM

Imagine you are to jump from an aircraft at an altitude of 1000 metres. You want to free-fall for 600 metres, knowing that in successive seconds you fall 5, 15, 25, 35, ... metres. How many seconds do you count before you pull the rip-cord?

justin
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1 Answers1

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Let $d_n$ be the total distance fallen after $n$ seconds. Looking at the pattern, we see that $d_1 = 5$, $d_2 = 20 = 5 + 15 = d_1 + (10 \cdot 2 - 5)$, $d_3 = 45 = 20 + 25 = d_2 + (10 \cdot 3 - 5)$, and for all $n > 1$: $$ d_n = d_{n-1} + (10n - 5) $$ Let's see what happens as we replace $n$ with $n-1$ successively: \begin{align*} d_n - d_{n-1} &= 10n - 5 \\ d_{n-1} - d_{n-2} &= 10(n - 1) - 5 \\ d_{n-2} - d_{n-3} &= 10(n - 2) - 5 \\ d_{n-3} - d_{n-4} &= 10(n - 3) - 5 \\ &~~\vdots \\ d_{2} - d_{1} &= 10(2) - 5 \\ \end{align*} Summing the $n - 1$ equations together, notice that the LHS telescopes, yielding: $$ d_n - d_1 = \sum_{k=2}^n(10k - 5) $$ The RHS is just an arithmetic series of $N = n - 1$ terms with initial term $a = 15$ and last term $L = 10n - 5$, so we obtain: \begin{align*} d_n - 5 &= \frac{n - 1}{2}(15 + (10n - 5)) \\ d_n &= \frac{n - 1}{2}(10(n + 1)) + 5 \\ d_n &= 5(n^2 - 1) + 5 \\ d_n &= 5n^2 \\ \end{align*} Letting $d_n = 600$, we conclude that the rip-cord should be pulled at $n = \sqrt{\frac{600}{5}} = 2\sqrt{30} \approx 10.95$ seconds.

Adriano
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  • That's a great answer.But could you explain the term 'successive seconds' in the context of this question. – justin Nov 13 '14 at 08:23
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    It means: in the first second, you fall $5$ metres; in the second second, you fall an additional $15$ metres; in the third second, you fall an additional $25$ metres; and so on. – Adriano Nov 13 '14 at 08:25
  • But can we predict pratically that within a second a person would fall 5 metres.I'm asking this to know whether there any applications related to finite differences. – justin Nov 13 '14 at 08:39
  • Yes. If you've taken physics, a more exact value is $\frac{1}{2}g \approx \frac{1}{2}(9.8) = 4.9$. – Adriano Nov 13 '14 at 08:42
  • okay.But is there any applications related to finite difference for field of computer science. – justin Nov 13 '14 at 08:45