Question

(a) Derive the formula s = ut + 1/2 at2, where the symbols have usual meanings.

Answer 1

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

is proved.

Explanation:

As we know that,

$average \: speed = \frac{distance \: covered}{time \: taken}$

$average \: speed = \frac{s}{t}$

And the average speed is also written as,

$average \: speed = \frac{u + v}{2}$

So, from equating this equation of average speed we get

$\frac{s}{t} = \frac{u + v}{2}$

From the 1st law of motion we get

$v = u + at$

Now,

$\frac{s}{t} = \frac{u + u + at}{2}$

$s = t \times ( \frac{2u + at}{2} )$

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

Hence, the 2nd law of motion is proved.

Answer 2

Draw Velocity-Time Graph

Explanation:

• For the derivation of Motion Equation - Velocity Time Graph is Drawn (Refer Figure Attachment)

• Let a body travelling a distance 's' in Time 't'.

The area of the space between the velocity-time graph represents distance travelled.

Area of OABC, AB & OC (time axis)

Hence,

• Distance travelled by body = Ar(OABC)

= Area of rectangle OADC + Area of ΔABD

In the figure, OADC is a rectangle and ΔABD.

(i) Area of rectangle OADC = $OA \times OC$

= $u \times t$

= ut  

(ii) Area of ΔABD = $(\frac {1}{2}) \times Area\ of\ rectangle\ AEBD$

= $(\frac {1}{2}) \times AD \times BD$

= $(\frac {1}{2}) \times t \times at$ (because AD = t and BD = at)

= $(\frac {1}{2}) at^2$

So, Distance travelled, s = Area of rectangle OADC + Area of triangle ABD

or $s = ut + (\frac {1}{2}) at^2$

This is the second equation of motion. Graphical Method can help with the derivation properly.