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Chapter 3: Problem 11

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

Expert verified
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

01

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.
02

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.
03

Expand the Total Displacement Equation

To prove this statement, we will expand the total displacement equation: \(\overrightarrow{\mathbf{D}} = (\overrightarrow{\mathbf{r}}_{2} - \overrightarrow{\mathbf{r}}_{1}) + (\overrightarrow{\mathbf{r}}_{3} - \overrightarrow{\mathbf{r}}_{2}) + ... + (\overrightarrow{\mathbf{r}}_{n} - \overrightarrow{\mathbf{r}}_{n-1})\)
04

Simplify the Total Displacement Equation

Notice that a lot of the terms cancel: \(\overrightarrow{\mathbf{D}} = \overrightarrow{\mathbf{r}}_{n} - \overrightarrow{\mathbf{r}}_{1}\)
05

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.

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Most popular questions from this chapter

Two displacement vectors each have magnitude \(20 \mathrm{km}\) One is directed \(60^{\circ}\) above the \(+x\) -axis; the other is directed \(60^{\circ}\) below the \(+x\) -axis. What is the vector sum of these two displacements? Use graph paper to find your answer.

An airplane has a velocity relative to the ground of \(210 \mathrm{m} / \mathrm{s}\) toward the east. The pilot measures his airspeed (the speed of the plane relative to the air) to be \(160 \mathrm{m} / \mathrm{s}\) What is the minimum wind velocity possible?
The range \(R\) of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at \(t=0\) with initial speed \(v_{i}\) at an angle \(\theta\) above the horizontal. (a) Find the time \(t\) at which the projectile returns to its original altitude. (b) Show that the range is \(R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}\) [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of \(\theta\) gives the maximum range? What is this maximum range?
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