Saved Bookmarks
| 1. |
Figute shows the motion of a particle along a straight line.Find the average velocity of the particle during the intervals (a)A to E, (b)B to E,(c )C to E, (d) D to E, ( e) C to D. |
Answer» SOLUTION : (a) As the particle moves from A to E, is the initial point and E is the final point. The slope of the line drawn from A to E i.e., `(Deltax)/(DELTAT)` gives the average velocity during that interval of time. The displacement `Deltax` is `X_(E)-X_(A)`=10cm-0cm=+10cm The time interval `Deltat_(EA)=t_(E)-t_(A)=10S`. `therefore` During this interval average velocity `barV=(Deltax)/(Deltat)=(+10cm)/(10S)=+1CMS^(-1)` (b)During the interval B to E. the displacement `Deltax=x_(E)-x_(B)=10cm-4cm=6cm` and `Deltat=t_(E)-t_(B)=10s-3S=7s.` `Therefore` Average velocity `barV=(Deltax)/(Deltat)=(6cm)/(7s)` =+0.857 `cms^(-1)=0.86cms^(-1)` (c) During the interval C to E,the displacement `Deltax=X_(E) -X_(c)=10cm0-12cm=2cm` nd `Deltat=t_(E)-t_(c)=10s-5s=5s` `therefore barv=(Deltax)/(Deltat)=(-2cm)/(5s)=-0.4 cms^(-1)` (d)During the interval D to E, the displacement `Deltax=x_(E)-x_(D)=10cm-12cm=-2cm` and the time interval `Deltat=t-(E)-t_(D)`=10s-8s=2 `therefore barv=(v)=(Deltax)/(deltat)=(-2cm)/(2s)=-1cms^(-1)` (e)During the interval C to D.the displacement `Deltax-x_(D)-x_(c)`=12cm 12 cm=0 and the time interval `Deltat=t_(D)-t_(c)`=8s-5s=3s `therefore` The average velocity `barv=(Deltax)/(Deltat)=(0m)/(3s)=0 ms^(-1)` (The particle has reached the same position during these 3s.The average velocity is zero because the displacement is zero). |
|