/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q2P Question: Compute your average ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question:Compute your average velocity in following two cases: (a) You walk 73.2 mat a speed of 1.22 m/sand then run 73.2at a speed of 3.05m/salong a straight track. (b) You walk for1 minat a speed of 1.22m/sand then run for1 min at3.05m/salong a straight track. (c) Graph x versus t for both cases and indicate how the average velocity is found on the graph.

Short Answer

Expert verified

Answer

  1. Average velocity for the first case is 1.74 m/s .
  2. Average velocity for the second case is 2,14 m/s

Step by step solution

01

Given data

In the first case, the distance covered while walking,x1 =73.2 m .

In the first case, the speed while walking,v1 =1.22 m/s .

In the first case, the distance covered while running,x2 =73.2 m .

In the first case, the speed while running, v2 =3.05 m/s .

In the second case, the time for walking,t1 =1 min .

In the second case, the speed while walking,v1 =1.22 m/s .

In the second case, the time for running, t2 =1 min .

In the second case, the velocity while running, v2 =3.05 m/s .

02

Understanding the average velocity

Average velocity is the ratio of the total displacement to the total time taken to occur that displacement.

The expression for the average velocity is as follows:

vavg=∆x∆t (i)

Here,∆x is the displacement, and ∆tis the time interval.

03

(a) Determination of the average velocity in the first case

The total time taken to cover the whole distance is calculated as follows:

∆t=x1v1+x2v2=73.2m1.22m/s+73.2m3.05m/s=60s+24s=84s

The total distance covered is calculated as follows:

∆x=x1+x2=73.2+73.2=146.4m

Now, the average velocity is as follows:

role="math" localid="1651662453157" ∆vavg=∆x∆t=146.4m84s=1.74m/s

Thus, the average velocity in the first case is 1.74 m/s.

04

(b) Determination of the average velocity in the second case

In this case, the total distance covered is calculated as follows:

∆x=v1×t1+v2×t2=1.22m/s×60s+3.05m/s×60s=73.2+183=256.2m

The total time interval is as follows:

∆t=60+60=120s

So, the average velocity is as follows:

∆vavg=∆x∆t=256.2m120s=2.135m/s

Therefore, the average velocity in the second case is 2.135 m/s.

05

(c) Determination of average velocity using an x-t graph 

You can find the average velocity using a graph ofx versus t.

Case (a)

The graph shows two different displacement paths with a blue line. The first one is the displacement while walking, and the second one is the displacement while running. If you take the slope of each line, you will get the velocity of walking and running. If you take the slope of the line connecting the initial displacement to the final displacement (black line), you will get the average velocity.

Case (b)

The graph shows two displacements. The first one is walking for,and the second one is running for. The slope of each line will give us the velocity of walking and running.If you take the slope of the line connecting the initial displacement to the final displacement (black line), you will get the average velocity.

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 91Ó°ÊÓ!

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

An electric vehicle starts from rest and accelerates at a rate of2.0m/s2in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of1.0m/s2until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?

A particle starts from the origin at t=0and moves along the positive xaxis. A graph of the velocity of the particle as a function of the time is shown in Fig. 2-46; the v-axisscale is set by vs 4.0m/s. (a) What is the coordinate of the particle at t=5.0s? (b) What is the velocity of the particle att=5.0s? (c) What the acceleration of the particle at t=5.0s? (d) What is the average velocity of the particle between t=1.0sand t=5.0s? (e) What is the average acceleration of the particle between t=1.0sand t=5.0s?

From t=0 tot=5.00min, a man stands still, and fromt=5.00mintot=10.0min, he walks briskly in a straight line at a constant speed of2.20m/s. What are a) his average velocityVavgb)his average acceleration aavgin time interval2.00minto8.00min? What are c)Vavgd)aavgin time interval3.00minto9.00min? e) Sketch x vs t and v vs t and indicate how the answers of (a) through (d) can be obtained from the graphs.

A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity that is 12mfrom the point of impact when the key is released. What is the speed of the boat?

An abrupt slowdown in concentrated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in traffic direction) or upstream, or it can be stationary. Figure 2-25shows a uniformly spaced line of cars moving at speed V=25 m/stoward a uniformly spaced line of slow cars moving at speedvs=5m/s. Assume that each faster car adds length L=12.0 m(car length plus buffer zone) to the line of slow cars when it joins the line, and assume it slows abruptly at the last instant. (a) For what separation distance d between the faster cars does the shock wave remain stationary? If the separation is twice the amount, (b) What is the Speed? (c) What is the Direction (upstream or downstream) of the shock wave?

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.