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Chapter 3: Q6CQ (page 119)

Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A+B the sum of the lengths of the two steps?

Short Answer

Expert verified

Two non-zero displacements can be zero only if they have equal magnitude but are headed in the opposite directions. Yes, the maximum distance we can end up from the starting point is the sum of the lengths of the two steps.

Step by step solution

01

Distance and displacement

A distance is the actual path covered by a moving object. It is a scalar quantity that cannot be zero or negative.

A displacement is the shortest path covered by a moving body from an initial position to reach a final position. It is a vector quantity, which can be zero or negative under certain conditions.

02

Ending up at the starting point

Two non-zero vectors add up to zero only if their magnitude is equal, but they are headed in the opposite directions.

Hence, if we take one step along the north and another step of the same length along the south, we will end up at the starting point.

03

Maximum distance

A distance is a scalar quantity, which can be added according to simple algebraic rules.

If you take one step of length A and another step of length B, then the maximum distance traveled is:

d=A+B

Hence, the maximum distance one can end up from the starting point is the sum of the lengths of the two steps.

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Most popular questions from this chapter

A cloud of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers.

For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

You fly \(32.0{\rm{ km}}\) in a straight line in still air in the direction \(35.0^\circ \) south of west.

(a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.)

(b) Find the distances you would have to fly first in a direction \(45.0^\circ \) south of west and then in a direction \(45.0^\circ \) west of north. These are the components of the displacement along a different set of axes鈥攐ne rotated \(45.0^\circ \).

(a) Repeat the problem two problems prior, but for the second leg you walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) north of east (which is equivalent to subtracting \({\rm{B}}\) from \({\rm{A}}\) 鈥攖hat is, to finding \({\rm{R'}} = {\rm{A}} - {\rm{B}}\)).

(b) Repeat the problem two problems prior, but now you first walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) south of west and then \(12.0{\rm{ m}}\) in a direction \(20.0^\circ \) east of south (which is equivalent to subtracting \({\rm{A}}\) from \({\rm{B}}\) 鈥攖hat is, to finding \({\rm{R''}} = {\rm{B}} - {\rm{A}} = - {\rm{R'}}\)). Show that this is the case.

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50km45.0掳north of west; then 4.70km60.0掳south of east; then 1.30km25.0掳south of west; then 5.10kmstraight east; then 1.70km5.00掳east of north; then 7.20km 55.0掳south of west; and finally 2.80km10.0掳north of east. What is his final position relative to the island?
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