Question

Prove that the displacement for a trip is equal to the vector sum of the displacements for each leg of the trip. [Hint: Imagine a trip that consists of \(n\) segments. The trip starts at position \(\overrightarrow{\mathbf{r}}_{1},\) proceeds to \(\overrightarrow{\mathbf{r}}_{2},\) then to \(\overrightarrow{\mathbf{r}}_{3}, \ldots\) then to \(\overrightarrow{\mathbf{r}}_{n-1},\) then finally to \(\overrightarrow{\mathbf{r}}_{n} .\) Write an expression for each displacement as the difference of two position vectors and then add them.]

Short answer

Question: Prove that the total displacement for a trip with multiple legs is the vector sum of the displacements for each leg. Answer: The total displacement for a trip with multiple legs is equal to the vector sum of the displacements for each leg, as shown by the simplified expression \(\overrightarrow{\mathbf{D}} = \overrightarrow{\mathbf{r}}_{n} - \overrightarrow{\mathbf{r}}_{1}\), which represents the total displacement for the trip.

Step-by-step solution

Define Displacement Vectors

For a trip with n segments, we define the displacement vector for each leg as follows: \(\overrightarrow{\mathbf{d}}_{i} = \overrightarrow{\mathbf{r}}_{i+1} - \overrightarrow{\mathbf{r}}_{i}\) for \(i = 1, 2, ..., n-1\), where \(\overrightarrow{\mathbf{r}}_{i}\) and \(\overrightarrow{\mathbf{r}}_{i+1}\) represent the starting and ending positions of the i-th leg, respectively, and \(\overrightarrow{\mathbf{d}}_{i}\) represents the displacement vector of the i-th leg.

Calculate Total Displacement

To calculate the total displacement, we will add up the individual displacement vectors: \(\overrightarrow{\mathbf{D}} = \sum_{i=1}^{n-1} \overrightarrow{\mathbf{d}}_{i}\) This is the same as performing a vector sum of all the individual displacement vectors.

Expand the Total Displacement Equation

To prove this statement, we will expand the total displacement equation: \(\overrightarrow{\mathbf{D}} = (\overrightarrow{\mathbf{r}}_{2} - \overrightarrow{\mathbf{r}}_{1}) + (\overrightarrow{\mathbf{r}}_{3} - \overrightarrow{\mathbf{r}}_{2}) + ... + (\overrightarrow{\mathbf{r}}_{n} - \overrightarrow{\mathbf{r}}_{n-1})\)

Simplify the Total Displacement Equation

Notice that a lot of the terms cancel: \(\overrightarrow{\mathbf{D}} = \overrightarrow{\mathbf{r}}_{n} - \overrightarrow{\mathbf{r}}_{1}\)

Conclusion

Thus, the total displacement for a trip with multiple legs is indeed equal to the vector sum of the displacements for each leg, as the expression for the total displacement \(\overrightarrow{\mathbf{D}}\) reduces to the difference between the final and initial positions, which represents the total displacement for the trip.