IIT JEE , AIEEE Forum - Physics , Mechanics

P90

this is a question from dc pandey mechanics volume-1. pg 52.

The acceleration of a particle moving in a straight line varies with displacement as a=(2*s).Find the corresponding velocity-displacement equation.

After carrying out necessary integration the final ans obtained is v²=2s²

^{This can also be xpressed as v=(+or-)[root(2)]*s. pleeeeeeeeeez explain each of the above 2 answers obtained.pleeeeeeeeeeeeeeeeeeez helppppppppppppppp!!!!!!!!!!!!!!!!!!!!!!}

^{i wud be realllllllly reallllllly grateful if u xplain in deeeeeeeeetail}

rooney

Hi. This is SHM happening here.So, for a given value of x you are getting two values of v which are opposite only in direction and not in magnitude. So suppose mean position is x=0. and extreme positions on both sides is x=A and x= - A. The particle will oscillate between x= - A and x=A. So, at a given position of x say x=A/2 , it will pass through this point twice. One when its going from x=0 to x=A and another when it is returning from x=A to x=0. So, these 2 velocities are opposite in sign but same in magnitude.

Hope its clear now.

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