Question
Establish graphically :
- v = u + at,
- s = ut + 1⁄2 at2,
- 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