A particle moving along x-axis has acceleration f, at time t, given by \(f = f_o[1-\frac{t}{T}]\) , where f0 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 particle’s velocity (vx ) is :
Answer & explanation
Correct answer: option 3
Acceleration : \(f = f_o[1-\frac{t}{T}] = \frac{dv}{dt}\)
integrating : \(\int dv = \int f_o[1-\frac{t}{T}] dt\)
\( v = f_o t - \frac{t^2}{2T} + C \)
Initial condition : v(t = o) = 0
\( 0 = 0 - 0 + C\) ⇒ C = 0
Thus, velocity at any time t : \( v(t) = f_o t - \frac{t^2}{2T} \)
when f = 0 ⇒ \(0 = f_o[1-\frac{t}{T}] \)
Thus, t = T
\( v(t) = f_o T - \frac{T^2}{2T} = \frac{f_0}{2T}\)