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# A particle moves in a straight line with the velocity as shown in the figure. At t=0,x=−16 m, then, A
The maximum value of the position coordinate of the particle is 54 m
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B
The maximum value of the position coordination of the particle is 36 m
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C
The particle is at the position of 36 m at t=18 s
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D
The particle is at the position of 36 m at t=30 s
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Solution

## The correct options are A The maximum value of the position coordinate of the particle is 54 m C The particle is at the position of 36 m at t=18 s D The particle is at the position of 36 m at t=30 sMaximum value of position coordinate = initial coordinate + area under the graph upto t=24 s (As upto t=24 s, the displacement of the particle will be positive ) xmax=−16+[(2×10)+(2+62)×(18−10)+1×62×(24−18)] =−16+[20+32+18]=54 m At time t=18 s xt=18=−16+Area of graph upto t=18 s xt=18=−16+[20+32]=36 m At time t=30 s xt=30=−16+Area of graph upto t=30 s xt=30=−16+[70−12×6×6]=36 m  Suggest Corrections  0      Similar questions
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