The position of a particle as it moves along a yaxis is given by y=2.0cmsinπt/4, with tin seconds and yin centimeters. (a) What is the average velocity of the particle between t=0and t=2.0s? (b) What is the instantaneous velocity of the particle at t=0,1.0, and t=2.0s?? (c) What is the average acceleration of the particle between t=0and t=2.0s? (d) What is the instantaneous acceleration of the particle at t=0,1.0, and 2.0s?

Short Answer

Expert verified
  1. The average velocity of the particle between t=0and t=2.0sis 1.0cm/s.

  2. The instantaneous velocity of the particle at t=2.0sis 1.6cm/s, 1.1cm/sand 0cm/srespectively.

  3. The average acceleration of the particle between t=0and t=2.0sis .0.79cm/s2.

  4. The instantaneous acceleration of the particle at t=0, t=1.0sand 2.0sis 0cm/s2,-0.87cm/s2and role="math" localid="1654751382053" -1.2cm/s2respectively.

Step by step solution

01

Given data

The position of the particle, y=(2.0cm)sin(πt4)

02

Understanding the average velocity, instantaneous velocity, average acceleration and instantaneous acceleration

The average velocity of an object is the ratio of total displacement to the total time.

Instantaneous velocity is the time rate of change of displacement at the given instant

Average acceleration is the change in velocity for a particular time interval whereas instantaneous acceleration is the time rate of change of velocity at a given instant of time.

The expression for the average velocity is given as follows:

vavg=ΔxΔt … (i)

Here, Δxis the net displacement and Δtis the time interval.

The expression for the instantaneous velocity is given as follows:

v=dxdt … (ii)

The expression for the average acceleration is given as follows:

aavg=ΔvΔt … (iii)

The expression for the instantaneous acceleration is given as follows:

a=dvdt … (iv)

03

(a) Determination of the average velocity between t = 0 to t = 2.0 s

The position of the particle at t=0sis,

y=(2.0cm)sin(π×04)=0cm

The position of the particle at t=2sis,

y=(2.0cm)sin(π×24)=(2.0cm)sin(π2)=2.0cm

Using equation (i), the average velocity is calculated as follows:

vavg=2.0cm-0cm2s-0s=1.0cm/s

Thus, the average velocity of the particle between t=0and t=2.0sis role="math" localid="1654752865421" 1.0cm/s.

04

(b) Determination of  the instantaneous velocity

Using equation (ii), the instantaneous velocity is,

v=ddt(2.0cmsinπt4)v=(π2cm)cos(πt4)

At t=0sthe instantaneous velocity is,

v=(π2)cos(0)=1.57cm/s1.6cm/s

Atthe instantaneous velocity is,

role="math" localid="1654753296911" v=(π2)cos(πt2)=0cm/s

At t=2sthe instantaneous velocity is,

Therefore, the instantaneous velocity of the particle at t=o,t=1.0sandt=2.0sis 1.6cm/s, 1.1cm/sand 0cm/srespectively.

05

(c) Determination of the average acceleration between  to 

Using equation (iii), the average acceleration is calculated as follows:

aavg=v(2.0s)-v(0s)Δt=0cm/s2-1.57cm/s22.0s-0s=-0.79cm/s2

Thus, the average acceleration of the particle betweent=0 and t=2.0sis-0.79cm/s2

06

(d) Determination of the instantaneous acceleration

Using equation (iv), the instantaneous acceleration is,

a=ddt{π2cmsinπ4}a=-(π28cm)sin(π4)

Att= 0 s, the instantaneous acceleration is,

a=-(π28cm)sin(0)=0cm/s2

Att= 1 s, the instantaneous acceleration is,

a=-(π28cm)sin(π4)=-0.87cm/s2

Att= 2 s, the instantaneous acceleration is,

a=-(π28cm)sin(π2)=-=1.2cm/s2

Thus, the instantaneous acceleration of the particle at t=0, t=1.0sand 2.0sis 0cm/s2,-0.87cm/s2and -=1.2cm/s2respectively.

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

At a construction site, a pipe wrench stuck the ground with a speed of 24 m/s(a)From what height was it inadvertently dropped? (b)How long was it falling? (c)Sketch graphs of y,v and a versus t for the wrench.

In 1889, at Jubbulpore, India, a tug-of-war was finally won after role="math" localid="1654754892705" 2h41min, with the winning team displacing the center of the rope 3.7m. In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.29m/s2 . (a)How long does such a car, initially travelling at 24.6 m/s, take to stop? (b)How far does it travel in this time? (c)Graph x vs t and v vs t for the deceleration.

A red train traveling at 72 km/hand a green train traveling at 144km/hare headed toward each other along a straight, level track. When they are 950 mapart, each engineer sees the other’s train and applies the brakes. The brakes slow each train at the rate of1.0m/s2 . Is there a collision? If so, answer yes, give the speed of the red train and the speed of the green train at impact, respectively. If not, answer no and give the separation between the trains when they stop.

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.

Sign-up for free