JEE Main & Advanced Physics One Dimensional Motion Equation of Kinematics

Equation of Kinematics

Category : JEE Main & Advanced

These are the various relations between \[u,\,\,\upsilon ,\,\,a,\,\,t\] and s for the particle moving with uniform acceleration where the notations are used as :

u = Initial velocity of the particle at time t = 0 sec

\[\upsilon =\] Final velocity at time t sec

a = Acceleration of the particle

s = Distance travelled in time t sec

\[{{s}_{n}}=\] Distance travelled by the body in nth  sec

(1) When particle moves with zero acceleration

(i) It is a unidirectional motion with constant speed.

(ii) Magnitude of displacement is always equal to the distance travelled.

(iii) \[\upsilon =u,\],  s = u t   [As a = 0]

(2) When particle moves with constant acceleration

(i) Acceleration is said to be constant when both the magnitude and direction of acceleration remain constant.

(ii) There will be one dimensional motion if initial velocity and acceleration are parallel or anti-parallel to each other.

(iii) Equations of motion                                         Equation of motion    

 (in scalar from)                                                        (in vector from)

\[\upsilon =u+at\]                                                    \[\vec{v}=\vec{u}+\vec{a}t\]     

\[s=ut+\frac{1}{2}a{{t}^{2}}\]                                  \[\vec{s}=\vec{u}t+\frac{1}{2}\vec{a}{{t}^{2}}\]     

\[{{\upsilon }^{2}}={{u}^{2}}+2as\]                         \[\vec{v}.\vec{v}-\vec{u}.\vec{u}=2\vec{a}.\vec{s}\]     

\[s=\left( \frac{u+v}{2} \right)\,t\]                            \[\vec{s}=\frac{1}{2}(\vec{u}+\vec{v})\,t\]                

\[{{s}_{n}}=u+\frac{a}{2}\,(2n-1)\]                                 \[{{\vec{s}}_{n}}=\vec{u}+\frac{{\vec{a}}}{2}\,(2n-1)\]


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