Establish graphically : (i) v = u + at, (ii) s = ut + 1⁄2 at², (iii) v² = u² + 2as.

Question

Establish graphically :

1. v = u + at,
2. s = ut + 1⁄2 at²,
3. v² = u² + 2as.

Answer

(i) Derivation of v = u + at

Change in velocity in time interval t = BE = BD – ED

If AE be drawn parallel to OD, then from the graph,

BD = BE + ED = BD + OA

or, v = BE + u

or, BE = v – u

Now, acceleration, a = Change\ in\ velocity\over time

= BE\over AE

= BE\over OD

Putting OD = t  ⇒ a = BE\over t

⇒  BE = at = v – u

therefore,v = u + at

(ii) Derivation of s = ut + 1⁄2 at²

In the figure, the distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABD.

Thus, Distance travelled = Area of the trapezium OABD

But, Area of the figure OABD

= Area of rectangle OAED + Area of triangle ABE

= Area of rectangle OAED + area of triangle ABE

Now, find out the area of rectangle OAED and the area of triangle ABE.

(1) Area of rectangle OAED

= (OA)×(OC)

= (u)×(t)

(2) Area of triangle ABE

= 1\over 2 AE×BE

= 1\over 2 t×at

= 1\over 2 at²

Distance travelled (s) is,

So, s = Area of rectangle OAED + Area of triangle ABE

= ut + 1\over 2 at²

(iii) Derivation of v² = u² + 2as.

In the given figure, the distance travelled (s)  by a body in time  (t) is given by the area of the figure OABC which is a trapezium.

Distance travelld = Area of the trapezium OABC

So, Area of trapezium OABC

= 1\over 2 ×sum of parallel sides×height

= 1\over 2 (OA + CB) × OC

Now, (OA + CB) = u + v and (OC) = t.

Putting these values in the above relation, we get:

s = (u+v)\over 2 t —- (1)

Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
v = u + at

So, t = (v-u)\over a

Now, put this value of t in equation (1), we get:

s = (v+u)(v-u)\over 2a

On further simplification,

2as = v² – u²

Finally the third equation of motion.

v² = u²+2as