Q124 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics

A particle moving along $X$-axis has acceleration $f$ at time $t$ given by $f=f_0\left(1-\frac{t}{T}\right)$, 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 is

  • A
    $\frac{1}{2} f_0 T^2$
  • B
    $f_0 T^2$
  • Correct answer
    C
    $\frac{1}{2} f_0 T$
  • D
    $f_0 T$

Solution

Acceleration of the particle is given as $f=f_0\left(1-\frac{t}{T}\right)$ $\ldots(\mathrm{i})$ $\therefore$ Velocity of particle is given as, $v=\int f d t=\int f_0\left(1-\frac{t}{T}\right) d t=f_0\left(t-\frac{t^2}{2 T}\right)+c$ $\ldots(\mathrm{ii})$ Given that, at $t=0, v=0$ Hence, from Eq. (i), we get $0=f_0(0-0)+c \Rightarrow c=0$ $\therefore$ From Eq. (ii), we have $v=f_0\left(t-\frac{t^2}{2 T}\right)$ $\ldots(\mathrm{iii})$ Suppose, acceleration of the particle becomes zero at time $t=t_1$. Hence, from Eq. (i), $0=f_0\left(1-\frac{t_1}{T}\right) \Rightarrow 1-\frac{t_1}{T}=0 \Rightarrow t_1=T$ Hence, velocity of particle at $t=t_1=T$, from Eq. (iii), $v=f_0\left(t_1-\frac{t_1^2}{2 T}\right)$ $\Rightarrow \quad v=f_0\left(T-\frac{T^2}{2 T}\right)=f_0\left(T-\frac{T}{2}\right)=\frac{f_0 T}{2}$

Hence, the correct answer is option: 'C'.

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