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?

Short Answer

Expert verified
  1. The particle is at x=15mat5.00s
  2. Velocity of the particle at5.00s is2m/s
  3. Acceleration of the particle at 5.00secis-2m/s2
  4. Average velocity between t=1stot=5s is3.5m/s
  5. Average acceleration of the particle between t=1stot=5s is0m/s2

Step by step solution

01

To understand the concept

The problem is based on the simple concept of calculating area of triangle. The calculation of region bounded by the curve within the given graph is called as area under the curve.

Formula:

The area of triangle is given by,

A=12base×height

02

(a): Determination of position of particle at t=5.0 s

To find the coordinates at 5sec, we need to know the displacement of the particle at 5sec. The displacement of the particle would be equal to the area under the plot at t=5sec.

Area of the triangle fort=2secis

A1=12×2×4=4m

Area of the rectangle fort=2sectot=4secis

A2=2×4=8m

Area of the shape fort=4sectot=5sec

A3=12×1×2+1×2=3m

So, the total displacement is

A1+A2+A3

4m+8m+3m=15m

So, the coordinates at localid="1663063994425" t=5secare5,15.

03

(b): Determination of velocity on of particle at t=5.0 s

Velocity at 5.00secis given from graph as 2m/s.

04

(c): Determination of acceleration of particle at t=5.0 s

Acceleration at 5.0seccan be found by finding the slope of the graph at t=5.0sec. It is equal to a=-2m/s2. It is negative since the velocity is reducing.

05

(d): Determination of average velocity of particle between t=1.0 s and t=5.0 s                 

Using the method used in (a), we can find the displacement at t=1sec, which is equal to1m, we already know the displacement att=5secwhich is15m

vavg=ΔxΔt=15-15-1=144=3.5m/s

So, the average velocity would be3.5m/s.

06

(e): Determination of average acceleration of particle between t=1.0 s and t=5.0 s        

The average acceleration at1.0sec and5.0sec

Velocity at t=1secis2m/sand velocity att=5secis2m/s

aavg=ΔvΔt=2-24=0

So, the average acceleration would be0.

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