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A particle of mass m moves on the x-axis as follows: It starts from rest at r= 0 from the pointr=0, and come to rest at 1 at the point x = 1. The velocity changes lincarly with time. No other information is available about its motion at intermediate times [ 0 lt t lt 1] . If alpha denotes the instantaneous acceleration of the particle , then |
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Answer» `alpha` cannot remain positive for all t in the interval 0 to 1 S= Area under velocity -time graph `V_("MAX")=(1)/(2)xxV_("max")xxt_(2)`  `IMPLIES V_("max") =(2x)/(t_(2))=2 m//s` `:. alpha =(V_("max"-4))/(t_(2)/(2))=(2-0)/((1)/(2))=4 m//s^(2)` `:.`In some post or points `|alpha| gt 4`. |
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