JEE Physics Kinematics Notes Set 02

Basic Terms

Distance is the path length (scalar); displacement is the shortest vector from start to end. Average speed = total distance/total time; average velocity = displacement/time. Instantaneous velocity \( v=\frac{dx}{dt} \) and acceleration \( a=\frac{dv}{dt}=v\frac{dv}{dx} \).

Equations of Uniformly Accelerated Motion

\( v=u+at \), \( s=ut+\tfrac12at^2 \), \( v^2=u^2+2as \), \( s_n=u+\tfrac a2(2n-1) \) (distance in the n-th second).

Motion Under Gravity

For a body thrown up with speed u: time of ascent \( u/g \), maximum height \( u^2/2g \), time of flight to the same level \( 2u/g \). A body dropped from rest covers distances in the ratio 1 : 3 : 5 in successive equal seconds.

Graphs

Slope of an x-t graph is velocity; slope of a v-t graph is acceleration; area under a v-t graph is displacement; area under an a-t graph is change in velocity. For constant acceleration, x-t is a parabola and v-t is a straight line.

Variable Acceleration

When acceleration is a function of time, integrate: \( v=u+\int a\,dt \), \( x=\int v\,dt \). When acceleration depends on position, use \( v\,dv=a\,dx \).

Worked Example

A body starts from rest with acceleration \( 2 m/s^2 \). Find the distance covered in the 5th second.
Solution: \( s_5=0+\frac22(2\times5-1)=9\ \text{m} \).