QUESTION IMAGE
Question
- start the marker at zero on the number line, and flip the coin. for heads, move the marker one unit right; for tails, move one unit left.
- conduct ten rounds of coin flips.
- in your evidence notebook, use vector arrow models to determine the displacement of your marker after ten coin flips.
analyze
how does the distance traveled by the coin compare to its displacement after ten flips? compare your results with those of other groups.
Step1: Define variables
Let \(x_i\) be the displacement in the \(i -\)th coin - flip. If it's heads, \(x_i = 1\); if it's tails, \(x_i=- 1\), where \(i = 1,2,\cdots,10\).
Step2: Calculate displacement
The displacement \(D\) after \(n = 10\) coin - flips is \(D=\sum_{i = 1}^{10}x_i\). Since each \(x_i\) is either \(1\) or \(-1\), the value of \(D\) depends on the number of heads \(h\) and tails \(t\) such that \(h + t=10\) and \(D=h - t=h-(10 - h)=2h - 10\). The possible values of \(D\) range from \(- 10\) (all tails) to \(10\) (all heads).
Step3: Calculate distance traveled
The distance traveled \(d\) is the sum of the absolute - values of each displacement. Since in each flip, the magnitude of the displacement is \(1\), the distance traveled \(d=\sum_{i = 1}^{10}|x_i|\). In each coin - flip, \(|x_i| = 1\), so \(d = 10\).
Step4: Compare distance and displacement
The distance traveled \(d = 10\) is always non - negative and represents the total length of the path covered. The displacement \(D\) is the net change in position from the starting point. The magnitude of the displacement \(|D|\leq10\). When \(D = 10\) (all heads) or \(D=-10\) (all tails), \(|D|=d = 10\). In other cases, \(|D|\lt d\).
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The distance traveled by the coin after ten flips is always \(10\) units. The displacement \(D\) ranges from \(-10\) to \(10\). The distance traveled is greater than or equal to the magnitude of the displacement. When all flips are heads or all flips are tails, the magnitude of the displacement is equal to the distance traveled; in other cases, the magnitude of the displacement is less than the distance traveled.