Saved Bookmarks
| 1. |
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, `(trianglex)/(trianglet)` gives the average velocity during that INTERVAL of time. The displacement `trianglex` is `x_(E)-x_(A)=10cm-0cm=+10cm` The time interval `trianglet_(EA)=t_(E)-t_(A)=10s`. During this interval average velocity `vecv=(trianglex)/(trianglet)=(+10cm)/(10s)=+1cms^(-1)` (b) During the interval B to E, the displacement `trianglex=x_(E)-x_(B)=10cm -4cm=6cm` and `trianglet=t_(E)-t_(B)=10s-3s=7s` Average VELOCTY `vecv=(trianglex)/(trianglet)=(6cm)/(7s)` `=+0.857 CMS^(-1) =0.86cms^(-1)` (c) During the interval C to E, the displacement `trianglex=x_(E)-x_(C)=10cm-12cm=2cm and ` `trianglet=t_(E)-t_(C)=10s-5s=5s` `vecv=(trianglex)/(trianglet)=(-2cm)/(5s)=-0.4cms^(-1)` (d) During the interval D to E, the displacement `trianglex=x_(E)-x_(D)=10cm-12cm=-2cm` and the time interval `trianglet=t_(E)-t_(D)=10s-8s=2s` ` vecv=(trianglex)/(trianglet)=(-2cm)/(2s)=-1cms^(-1)` (e) During the interval C to D, the displacement `trianglex-x_(D)-x_(C)=12cm ` 12cm=0 and the time interval. `triangle t=t_(D)-t_(C)=8s-5s=3s` The average velocity `vecv=(trianglex)/(trianglet)=(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). |
|