Explain why a vector cannot have a component greater than its own magnitude.

Short Answer

Expert verified

The vector cannot have a component greater than its own magnitude, because the sine and cosine components are always either less than or equal to one.

Step by step solution

01

Components of vector

Any vector can be resolved into two components. One component is along the horizontal direction and other component is along the vertical direction. The horizontal component of vector is the cosine component and the vertical component is the sine component.

02

Reason for magnitude of vector not being greater than its magnitude

Suppose the sine and cosine components of vector A are Asinθand Acos\(\theta \). The sine and cosine components are always less than or equal to one, so the magnitudes of components are either less than or equal to its own magnitude.

Therefore, the vector cannot have a component greater than its own magnitude, because the sine and cosine components are always either less than or equal to one.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg \({\rm{B}}\), which is \(20.0\;{\rm{m}}\) in a direction exactly \(40^\circ \) south of west, and then leg \({\rm{A}}\), which is \(12.0\;{\rm{m}}\) in a direction exactly \(20^\circ \) west of north. (This problem shows that \({\rm{A}} + {\rm{B}} = {\rm{B}} + {\rm{A}}\).)

If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?

(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a\(32^\circ \)ramp at a speed of\(40.0{\rm{ m}}/{\rm{s}}\)\(\left( {144{\rm{ km}}/{\rm{h}}} \right)\). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are\(20.0{\rm{ m}}\)long?

(b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

An archer shoots an arrow at a 75.0mdistant target; the bull’s-eye of the target is at same height as the release height of the arrow.

(a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems.

(b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch3.50m above the release height of the arrow. Will the arrow go over or under the branch?

Repeat Exercise using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B+A=A+B.) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free