Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

At 6 safter 3:00, a butterfly is observed leaving a flower whose location is<6,-3,10>mrelative to an origin on top of a nearby tree. The butterfly flies until 10 safter 3:00, when it alights on a different flower whose location is<6.8,-4.2,11.2>m relative to the same origin. What was the location of the butterfly at a time 8.5 safter 3:00? What assumption did you have to make in calculating this location?

Short Answer

Expert verified

The location of the butterfly at a time 8.5s is 6.5,-3.75,10.75m.

The assumption made in this question is that the average velocity is constant.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The initial location of butterfly is, 6,-3,10m.
  • Time at which the butterfly was at initial location is, 6 s.
  • The final location of the butterfly is, 6.8,-4.2,11.4m.
  • Time by which the butterfly reached to the final location is, 10 s.
02

 Step 2: Significance of Newtonโ€™s first law in calculating the distances

This law states that a body will continue to proceed at a uniform velocity if no external force acts on that body.

The equation of the average velocity and the displacement gives the location of the butterfly.

03

Determination of the average velocity of the butterfly

The equation of the average velocity of the butterfly can be expressed as follows,

vavg=r2-r1t2-t1

Here, vavg is the average velocity, r1 is the initial location of butterfly, r2 is the final location of the butterfly, t1 is the time taken by the butterfly to reach to the initial location and t2 is the time taken by butterfly to reach at the final location.

For data-custom-editor="chemistry" r1=6,-3,10m, data-custom-editor="chemistry" r2=6.8,-4.2,11.2m, data-custom-editor="chemistry" t1=6sand data-custom-editor="chemistry" t2=10s.

data-custom-editor="chemistry" vavg=6.8,-4.2,11.2m-6,-3,10m10s-6svavg=0.8,-1.2,1.2m4s=0.2,-0.3,0.3m/s

04

Determination of the location of the butterfly

The equation of displacement of the butterfly is expressed as follows,

s=s0+vavgt

Here, s is the displacement of the butterfly, s0 is the initial displacement of the butterfly, vavg is the average velocity of the butterfly, and t is the time taken by the butterfly to reach the final location.

For vavg=0.2,-0.3,0.3m/s,s0=6,-3,10m and t=10s-8.5s=1.5s.

s=6,-3,10m+0.2,-0.3,0.3m/sร—2.5s=6.5,-3.75,10.75m

Thus, the location of the butterfly at a time 8.5 s is 6.5,-3.75,10.75.

The assumption made in this question is that the average velocity is constant as the velocity may have changed with time. But here it is taken as constant.

Thus, the assumption made in this question is that the average velocity is constant.

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

Which of the following are vectors?

(a)rโ†’/2(b)|rโ†’|/2(c)<rx,ry,rz>(d)5ยทrโ†’

Question: A car moves along a straight road. It moves at a speed of 50 km/h for 4 minutes, then during 4 minutes it gradually speeds up to 100 km/h, continues at this speed for 4 minutes, then during 4 minutes gradually slows to a stop. Make a sketch like the figures in Section 1.2, marking dots for the position along the road every minute.

Why do we use a spaceship in outer space, far from other objects, to illustrate Newton's first law? Why not a car or a train? (More than one of the following statements may be correct.) (1) A car or train touches other objects, and interacts with them. (2) A car or train can't travel fast enough. (3) The spaceship has negligible interactions with other objects. (4) A car or train interacts gravitationally with the Earth. (5) A spaceship can never experience a gravitational force.

In which of these situations is it reasonable to use the approximate equation for the momentum of an object, instead of the full relativistically correct equation? (1) A car traveling on an interstate highway (2) A commercial jet airliner flying between New York and Seattle (3) A neutron traveling at 2700 meters per second (4) A proton in outer space traveling at 2ร—108 m/s (5) An electron in a television tube traveling3ร—106 m/s.

Here are the positions at three different times for a bee in flight (a beeโ€™s top speed is about 7 m/s ). Time6.3s6.8s7.3sPosition<-3.5,9.4,0>m<-1.3,6.2,0>m<0.5,1.7,0>m(a)Between 6.3 s and 6.8 s , what was the beeโ€™s average velocity? Be careful with signs.

(b)Between6.3 s and 7.3 s, what was the beeโ€™s average velocity? Be careful with signs.

(c)Of the two average velocities you calculated, which is the best estimate of the beeโ€™s instantaneous velocity at time 6.3 s ?

(d)Using the best information available, what was the displacement of the bee during the time interval from 6.3 s to 6.33 s ?

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