Question

The displacement y(t) of a particle depends on time according to equation y(t) =a1t-a2tsquare. What is the dimension of a1 and a2?

Answer 1

Explanation:

We know , v = dx/dt

We know , v = dx/dta = dv/dt

We know , v = dx/dta = dv/dtx = ao + a1t + a2t2

We know , v = dx/dta = dv/dtx = ao + a1t + a2t2v= dx/dt = 0 + a1 + 2a2t

We know , v = dx/dta = dv/dtx = ao + a1t + a2t2v= dx/dt = 0 + a1 + 2a2ta=dv/dt = d/dt (a1+2a2t)

We know , v = dx/dta = dv/dtx = ao + a1t + a2t2v= dx/dt = 0 + a1 + 2a2ta=dv/dt = d/dt (a1+2a2t)=0+2a2

We know , v = dx/dta = dv/dtx = ao + a1t + a2t2v= dx/dt = 0 + a1 + 2a2ta=dv/dt = d/dt (a1+2a2t)=0+2a2differntiation of constant is 0...

We know , v = dx/dta = dv/dtx = ao + a1t + a2t2v= dx/dt = 0 + a1 + 2a2ta=dv/dt = d/dt (a1+2a2t)=0+2a2differntiation of constant is 0...so accleration of particle = 2a2

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