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Study the following velocity-time graph....

Study the following velocity-time graph. Determine the displacement and distance in the time interval `t=0` to `t=12 s`. Sketch the acceleration-time graph.

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`{:("Time","Area"),(t=0 "to" t=2s,1/2xx6xx2=6),(t=2 "to" 4s,6xx2=12),(t=4 "to" 5s,1/2(6+12)xx1=9),(t=5 "to" 7s,1/2xx12xx2=12),(t=7 "to" 12s,1/2(2+5)xx4=14),(t=0 "to" t=12s,):}`
Displacement`=6+12+9+12-14=25 m`
Distance`=6+12+9+12+14=53 m`
a-t graph
`{:(t=0 "to" 2s,"Slope"=3),(t=2 "to" 4s,"Slope"=0),(t=4 "to" 5s,"Slope"=6),(t=5 "to" 7s,"Slope"=-6),(t=7 "to" 9s,"Slope"=-2),(t=9 "to" 11s,"Slope"=0),(t=11 "to" 12s,"Slope"=4):}`,
Slope of v-t graph gives acceleration.
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Knowledge Check

  • The velocity-time graph of an objects is as shown. The displacement during the interval 0 to t_(4) is

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    Speeds of particle at t=0 and t=2T are same.

  • The area under the acceleration-time (a-t) graph gives

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    Displacement
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