derive the equation of motion v=u+at by graphical method
Answers
TO DERIVE v = u + at BY GRAPHICAL METHOD
This is a graph of uniform acceleration with ‘u’ as initial velocity and ‘v’ as final velocity.
Initial velocity = u = OP
Final velocity = v = RN
= RQ + QN
v = u + QN …..(i)
Acceleration, a = slope of line PN
\displaystyle a=\frac{QN}{PQ}=\frac{QN}{t}
\displaystyle QN=at
…..(ii)
Putting the value of QN from equation (ii) into equation (i), we get v = u + at
TO DERIVE
\displaystyle s=ut+\frac{1}{2}a{{t}^{2}}
BY GRAPHICAL METHOD
In the above speed-time graph, the distance travelled is given by
Distance travelled = Area of figure OPNR
= Area of DPNQ + Area of rectangle OPQR
(1) Area of triangle PNQ =
\displaystyle \frac{1}{2}\times base\times height
=
\displaystyle \frac{1}{2}\times PQ\times NQ
\displaystyle =\frac{1}{2}\times t\times (v-u)
=
\displaystyle \frac{1}{2}\times t\times at
[As v = u + at and v – u = at]
Area of DPNQ =
\displaystyle \frac{1}{2}a{{t}^{2}}
(2) Area of rectangle OPQR = OP ´ PQ
= u ´ t
= ut
Distance travelled = Area of DPNQ + Area of rectangle OPQR
\displaystyle S=\frac{1}{2}a{{t}^{2}}+ut
\displaystyle S=ut+\frac{1}{2}a{{t}^{2}}
TO DERIVE
\displaystyle {{v}^{2}}-{{u}^{2}}=2aS
BY GRAPHICAL METHOD
In the above speed time graph distance travelled (S) = Area of trapezium OPNR
\displaystyle S=\frac{1}{2}\times
(sum of parallel sides) ´ height
\displaystyle S=\frac{1}{2}\times
(OP + RN) ´ OR
\displaystyle S=\frac{1}{2}\times
(u + v) ´ t
\displaystyle S=\frac{1}{2}\times
(v + u) t …(i)
But v = u + at
at = v – u
\displaystyle t=\left( \frac{v-u}{a} \right)
… (ii)
Putting this value of ‘t’ from equation (ii) into equation (i) we get that ___
\displaystyle S=\frac{1}{2}\times
(v + u)
\displaystyle \left( \frac{v-u}{a} \right)
\displaystyle S=\frac{\left( v+u \right)\left( v-u \right)}{2a}
\displaystyle S=\frac{{{v}^{2}}-{{u}^{2}}}{2a}
[As (a + b) (a – b) =
\displaystyle {{a}^{2}}-{{b}^{2}}
]
2as =
\displaystyle {{v}^{2}}-{{u}^{2}}
\displaystyle {{v}^{2}}-{{u}^{2}}=2as