Question

Derive:
$s = ut + \frac{1}{2} a {t}^{2}$
​

Answer 1

Explanation:

Or in simple way :-

___________________

Average velocity=u+v/2

or, s=distance

s=v^t

so. s= (u+v)t/2

v=u+at

so. s=(u+u+at)t/2

=(2u+at)t/2

=2ut+at sq.

s=ut+1/2at sq.

Answer 2

Answer:

To derive :

2nd Equation of motion using graphical methods.

s = ut + ½at²

Proof:

Lets consider a velocity-time graph as attached above.

Now we know that displacement is given by the area under the velocity time graph.

Let initial velocity be u , final velocity be v , acceleration be a , time be t and distance be s

Now, area under the curve:

∴ s = Area of the trapezium

=> s = ½ (sum of parallel sides) × dist. between them

=> s = ½ (u + v ) × t ......(1)

Now we know that :

v = u + at .....(from 1st eq. of motion) ......(2)

So putting value of "v" in eq.(1)

∴ s = ½ { u + (u + at)} × t

=> s = ½ { 2u + at } × t

=> s = ut + ½at²

[Hence proved .]