A particle moving along a straight line with uniform acceleration a. Let its velocity change from u to v in an interval of time t. Its displacement in this duration is s then

Given that

A particles moving along a straight line

Initial velocity = u

Final velocity = v

Uniform acceleration = a

Time taken = t

Then we get

i) Distance travelled s = Average velocity × Time

$\Rightarrow \mathrm{s}={\mathrm{v}}_{\mathrm{avg}}\times \mathrm{t}\Rightarrow \mathrm{s}=\left(\frac{\mathrm{u}+\mathrm{v}}{2}\right)\times \mathrm{t}$

$\Rightarrow \frac{\mathrm{s}}{\mathrm{t}}=\frac{\mathrm{u}+\mathrm{v}}{2}\because \mathrm{u}=\mathrm{v}-\mathrm{at}$

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\Rightarrow \frac{\mathrm{s}}{\mathrm{t}}=\frac{\mathrm{v}-\mathrm{at}+\mathrm{v}}{2}\text{ or }\frac{\mathrm{s}}{\mathrm{t}}=\frac{\mathrm{u}+\mathrm{u}+\mathrm{at}}{2}$

$\Rightarrow \frac{s}{t}=v-\frac{\text{ at }}{2},\Rightarrow \frac{s}{t}=u+\frac{\text{ at }}{2}$

We get all the options are true 

$\text{ i.e }s=v_{\mathrm{avg}}\times \mathrm{t},\frac{\mathrm{s}}{\mathrm{t}}=\left(\frac{\mathrm{u}+\mathrm{v}}{2}\right),\mathrm{s}=\mathrm{u}+\frac{\mathrm{at}}{2},\mathrm{s}=\left(\frac{\mathrm{u}+\mathrm{v}}{2}\right)\times \mathrm{t},\frac{\mathrm{s}}{\mathrm{t}}=\mathrm{v}-\frac{\mathrm{at}}{2}$