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CBSE Class 11-science Answered

The position (x) of a particle at time t, is given by the relation, x(t)=v0/a(1-e-at), where v0 is a constant and a>0. The dimensions of v0 and a are respectively  

Asked by Anil | 17 May, 2017, 09:21: PM
answered-by-expert Expert Answer
begin mathsize 12px style The space position space of space the space particle space at space straight a space time space straight t space is space given space as straight x left parenthesis straight t right parenthesis equals fraction numerator straight v subscript straight o over denominator straight a left parenthesis 1 minus straight e to the power of negative at end exponent right parenthesis end fraction As space straight x space denotes space position comma space left square bracket straight x right square bracket space equals space straight m comma space straight i. straight e. space distance space in space metre. Here space the space exponential comma space straight e to the power of at space must space be space dimensionless. Hence space left square bracket straight a right square bracket equals straight s to the power of negative 1 end exponent space or space inverse space time. or space straight a space equals space left square bracket straight M to the power of 0 space straight L to the power of 0 space straight T to the power of negative 1 end exponent space right square bracket rightwards arrow fraction numerator open square brackets straight V subscript straight o close square brackets over denominator open square brackets straight a close square brackets end fraction space equals space open square brackets straight V subscript straight o close square brackets space straight s space equals space straight m Hence space straight V subscript 0 space has space dimensions space of space velocity comma space straight i. straight e. space left square bracket straight V subscript 0 right square bracket equals ms to the power of negative 1 end exponent or space straight V subscript 0 space equals space left square bracket straight M to the power of 0 space straight L to the power of 1 space straight T to the power of negative 1 end exponent right square bracket end style
Answered by Yashvanti Jain | 18 May, 2017, 02:25: PM

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