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figure 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, A 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_(B)-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)` `=+0.857cms^(-1)=0.86cms-1` (c ) During the interval C to E, the displacement `Deltax=x_(E)-x_(C)=10cm-12cm=2cm` and `Deltat=t_(E)-t_(C)=10s-5s=5s` `therefore barv=(Deltax)/(Deltat)=(-2cm)/(5s)=-0.4cms^(-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=2s` `therefore barv=(Deltax)/(Deltat)=(-2cm)/(2s)=-1cms^(-1)` (e) During the interval C to D, the displacement `Deltax=x_(D)-x_(C)=12cm-12cm=0` and the time interval `Deltat=t_(D)-t_(C)=8s-5s=3s` `therefore` The average velocity `barv=(Deltax)/(Deltat)=(0m)/(3s)=0ms^(-1)` (The particle has reached the same position during these 3s. The average velocity is ZERO because the displacement is zero). |
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