A particle moving along x-axis has an acceleration f at time t, given by \(f = f_{0}(1-\frac{t}{T})\), where \(f_{0}\) and T are constants. The particle at T = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the velocity (va) of the particle is then:(a) \(\frac{1}{2}f_{0}T\)(b) \(f_{0}T\)(c) \(\frac{1}{2}f_{0}T^{2}\)(d) \(f_{0}T^{2}\)


[latex]f = f_{0}(1-\frac{t}{T})[/latex] where [latex]f_{0}[/latex] and T are constants

At t = 0 and v = 0

When f = 0 then,

0 = [latex]f_{0}(1-\frac{t}{T})[/latex]

T = t ———— (1)

f = (final – initial velocity)/ Time

[latex]\frac{dv}{dt}=f_{0}(1-\frac{t}{T})[/latex] [latex]\int dv=\int f_{0}(1-\frac{t}{T})dt[/latex] [latex]v=f_{0}(t-\frac{^{t^{2}}}{2T})[/latex]

From equation (1)

We know that, T = t


Therefore, the correct answer is option (a)

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