Question

Derive the third kinematic equation from the first two kinematic equations.

Answer 1

the first two kinematic equations are,

v=u+a*t ----(1)

s=u*t+1/2*a*t^2 ---(2)

u ----> initial velocity of the object is u

v ----> final velocity of the object is v

a ----> constant acceleration

s----> distance travelled

t ----> time taken

from equation (1),

t=v-u/a

from equation (2),

s=u*t+1/2*a*t^2

s=u*(v-u)/a+1/2*a*(v-u)^2/a^2

s=(u*v-u^2)/a+1/2*(v^2-u^2-2*u*v)/a

s=(u*v-u^2)/a+(v^2-u^2-2*u*v)/2a

s=2(u*v-u^2)/2a+(v^2-u^2-2*u*v)/2a

2*a*s=2(u*v-u^2)+(v^2-u^2-2*u*v)

2*a*s=2*u*v-2*u^2+v^2-u^2-2*u*v

2*a*s=v^2-u^2

====> v^2-u^2=2*a*s ----->is answer