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A particle moving along x-axis has acceleration $f$, at time $t$, given by $f = f_0\left(1 - \dfrac{t}{T}\right)$, 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 particle's velocity ($v_x$) is
A
$\dfrac{1}{2}f_0 T^2$
B
$f_0 T^2$
C
$\dfrac{1}{2}f_0 T$
D
$f_0 T$
Detailed Solution
Here $f = f_0\left(1 - \dfrac{t}{T}\right)$, i.e. $\dfrac{dv}{dt} = f_0\left(1 - \dfrac{t}{T}\right)$
$dv = f_0\left(1 - \dfrac{t}{T}\right)dt$
Integrating: $v = \int f_0\left(1 - \dfrac{t}{T}\right)dt = f_0\left(t - \dfrac{t^2}{2T}\right) + C$, where $C$ is the constant of integration.
At $t = 0$, $v = 0$: $0 = f_0(0 - 0) + C \Rightarrow C = 0$
$\therefore v = f_0\left(t - \dfrac{t^2}{2T}\right)$
If $f = 0$, then $0 = f_0\left(1 - \dfrac{t}{T}\right) \Rightarrow t = T$
Hence the particle's velocity gained in the time interval $t = 0$ to $t = T$ is $v_x = \int_{t=0}^{t=T} f_0\left(1 - \dfrac{t}{T}\right)dt = f_0\left[t - \dfrac{t^2}{2T}\right]_0^T$
$v_x = f_0\left(T - \dfrac{T^2}{2T}\right) = f_0\left(T - \dfrac{T}{2}\right)$
$v_x = \dfrac{1}{2}f_0 T$
$dv = f_0\left(1 - \dfrac{t}{T}\right)dt$
Integrating: $v = \int f_0\left(1 - \dfrac{t}{T}\right)dt = f_0\left(t - \dfrac{t^2}{2T}\right) + C$, where $C$ is the constant of integration.
At $t = 0$, $v = 0$: $0 = f_0(0 - 0) + C \Rightarrow C = 0$
$\therefore v = f_0\left(t - \dfrac{t^2}{2T}\right)$
If $f = 0$, then $0 = f_0\left(1 - \dfrac{t}{T}\right) \Rightarrow t = T$
Hence the particle's velocity gained in the time interval $t = 0$ to $t = T$ is $v_x = \int_{t=0}^{t=T} f_0\left(1 - \dfrac{t}{T}\right)dt = f_0\left[t - \dfrac{t^2}{2T}\right]_0^T$
$v_x = f_0\left(T - \dfrac{T^2}{2T}\right) = f_0\left(T - \dfrac{T}{2}\right)$
$v_x = \dfrac{1}{2}f_0 T$