The velocity of a particle is $v = v _{0}+ gt + Ft ^{2}$. Its position is $x=0$ at $t=0$; then its displacement after time $( t =1)$ is :

The velocity-time graph of a particle in one-dimensional motion is shown in Figure.

Which of the following formulae are correct for describing the motton of the particle over the time-interval $t_1$ to $t_2$

$(a)$ $x\left(t_{2}\right)=x\left(t_{1}\right)+v\left(t_{1}\right)\left(t_{2}-t_{1}\right)+(1 / 2) a\left(t_{2}-t_{1}\right)^{2}$

$(b)$ $v\left(t_{2}\right)=v\left(t_{1}\right)+a\left(t_{2}-t_{1}\right)$

$(c)$ $v_{\text {average}}=\left(x\left(t_{2}\right)-x\left(t_{1}\right)\right) /\left(t_{2}-t_{1}\right)$

$(d)$ $a_{\text {average}}=\left(v\left(t_{2}\right)-v\left(t_{1}\right)\right) /\left(t_{2}-t_{1}\right)$

$(e)$ $x\left(t_{2}\right)=x\left(t_{1}\right)+v_{\text {average}}\left(t_{2}-t_{1}\right)+(2 / 2) a_{\text {average}}\left(t_{2}-t_{1}\right)^{2}$

$(f)$ $x\left(t_{2}\right)-x\left(t_{1}\right)=$ area under the $v -t$ curve bounded by the $t -$axis and the dotted line shown.

The graph of displacement v/s time is given below Its corresponding velocity-time graph will be

The displacement of the particle varies with time according to the relation $x = \frac{k}{b}\,[1 – {e^{ – bt}}]$. Then the velocity of the particle is