Kinematic Equations for Uniformly Accelerated Motion

Class 11
Physics
Chapter 3
Motion In A Straight Line

1. KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION v=v +at x = ½ at 2 + V o t v 2 = V o 2 +2 ax

2. Acceleration= = a =   DERIVATION v=v +at Therefore at= V-V o V= V o + at V o - Intial velocity V- Final Velocity t- Time taken

3. Area under the Graph = Area of Rectangle ACDO + Area of Triangle ABC =L × B + ½ × l × b = AO × OD + ½ × AC × BC = V o × t + ½ × t × (V-V o ) As we have studied that the area under v-t curve represents the displacement. Therefore, the displacement x of the object is : x = ½ (V-V o )t + V t   But we know that , v − v = at Therefore, x = ½ at 2 + V o t DERIVATION x = ½ at 2 + V o t V o - Intial velocity V- Final Velocity t- Time taken

4. DERIVATION v 2 = V o 2 +2 ax From the Equation, x = ½ at 2 + V o t x= (½at + V o )t =( V o )t   From V=V o + at at= V-V o Taking LCM x=( x= ( )t   From V=V o + at t= V-V o /a x= ( )( ) 2ax= V 2 - V o 2   Therefore v 2 = V o 2 +2 ax V o - Intial velocity V- Final Velocity t- Time taken Area of Trapezium= ½ × Distance between parallel sides (Sum of parallel sides) x= ½ × OD (BD+AO) = ½× t( V+V o ) From V=V o + at t= V-V o /a x= ½ × ( ) ( V+V o ) From Algebraic Idnentity - (a-b)( a+b )= a 2 -b 2 X= ½ × 2ax=V 2 -V o 2   Therefore V 2= V o 2 +2ax METHOD 1 METHOD 2