A particle moves along a straight line and its velocity depends on time as $\mathrm{v}=4\mathrm{t}-{\mathrm{t}}^2$. Then for first 5 s:

Average velocity,

$\overrightarrow{\mathrm{v}}=\frac{{\int }_0^5 \mathrm{vdt}}{{\int }_0^5 \mathrm{dt}}=\frac{{\int }_0^5 \left(4\mathrm{t}-{\mathrm{t}}^2\right)\mathrm{dt}}{{\int }_0^5 \mathrm{dt}}$

$=\frac{{\left[2{\mathrm{t}}^2-\frac{{\mathrm{t}}^3}{3}\right]}_0^5}{5}=\frac{50-\frac{125}{3}}{5}=\frac{25}{3\times 5}=\frac{5}{3}$

For average speed, let us put v = 0, which gives t = 0 and t = 4 s

$\therefore \text{ Average speed }=\frac{\left|{\int }_0^4 \mathrm{vdt}\right|+\left|{\int }_4^5 \mathrm{vdt}\right|}{{\int }_0^5 \mathrm{dt}}=\frac{\left|{\int }_0^4 \left(4\mathrm{t}-{\mathrm{t}}^2\right)\mathrm{dt}\right|+\left|{\int }_4^5 \mathrm{vdt}\right|}{5}$

$=\frac{{\left[2{\mathrm{t}}^2-\frac{{\mathrm{t}}^3}{3}\right]}_0^4+{\left[2{\mathrm{t}}^2-\frac{{\mathrm{t}}^3}{3}\right]}_4^5}{5}$

$=\frac{\left|{\left[2{\mathrm{t}}^2-\frac{{\mathrm{t}}^3}{3}\right]}_0^4\right|+\left|{\left[2{\mathrm{t}}^2-\frac{{\mathrm{t}}^3}{3}\right]}_4^5\right|}{5}=\frac{13}{5} {\mathrm{ms}}^{-1}$

For acceleration : $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(4\mathrm{t}-{\mathrm{t}}^2\right)=4-2\mathrm{t}$

At t = 0, a = 4 ms^-2

Therefore, options (3) and (4) are correct, and options (1) and (2) are wrong.