20. A particle moving along x-axis has acceleration f, at time t, given by f = f0(1 − 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

20. A particle moving along x-axis has acceleration f, at time t, given by \(f = f0( 1 − \frac{t }{T} )\)  , 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 \(( v_x )\) is

(a)  \( \frac{1}{2}f_0 T^2 \)

(b)  \( f_0 T^2 \)

(c)  \( \frac{1}{2} f_0 T \)

(d)  \( f_0 T \)

Answer :  (c)  \( \frac{1}{2} f_0 T \)

Acceleration is changing with respect to time.

\[ a = f_0\!\left(1 - \frac{t}{T}\right) = 0 \]

\[ 1 - \frac{t}{T} = 0 \]

\[ t = T \]

\[ \frac{dv}{dt} = f_0\!\left(1 - \frac{t}{T}\right) \]

\[ \int_{\substack{v=0}}^{v} dv = \int_{\substack{t=0}}^{\substack{t=T}} f_0 \left(1 - \frac{t}{T}\right)\,dt \]

\[ \left[ v \right]_{v=0}^{v=v} = f_0 \int \left[ 1 - \frac{t}{T} \right]\,dt \]

\[ = f_0 \left[ t - \frac{t^2}{2T} \right]_{t=0}^{t=T} \]

\[ [ v - 0 ] = f_0 \left( [ T - \frac{T^2}{2T} ] - 0 \right) \]

\[ v = f_0 \left( T - \frac{T}{2} \right) \]

\[ v = \frac{f_0 T}{2} \]