Question

Show that for a uniformly accelerated motion starting from velocity u and acquiring velocity v has average velocity equal to arithmetic mean of the initial (u) and final velocity (v).

Answer 1

Answer:

(V + U)/2

Explanation:

Initial Velocity = U

Final Velocity = V

Uniform acceleration = a

Time Taken = (V - U)/a

Distance Covered S = ut + (1/2)at²

S = U(V - U)/a  + (1/2)a((V - U)/a)²

=>S = ((V - U)/a )(U + (V- U)/2)

=> S =  ((V - U)/a )(V + U)/2

Average Velocity = S /T

=> Average Velocity  =  ((V - U)/a )((V + U)/2) / ((V - U)/a)

= (V + U)/2

arithmetic mean of the initial (u) and final velocity (v). = (U + V)/2

(V + U)/2  =  (U + V)/2

Hence Proved

Answer 2

$\huge\bigstar\underline\mathcal\blue{Answer:-}$

Solution:

Initial Velocity = U

Final Velocity = V

Uniform acceleration = a

Time Taken = (V - U)/a

Distance Covered S = ut + (1/2)at²

S = U(V - U)/a  + (1/2)a((V - U)/a)²

=S = ((V - U)/a )(U + (V- U)/2)

= S =  ((V - U)/a )(V + U)/2

Average Velocity = S /T

= Average Velocity  =  ((V - U)/a )((V + U)/2) / ((V - U)/a)

= (V + U)/2

arithmetic mean of the initial (u) and final velocity (v). = (U + V)/2

(V + U)/2  =  (U + V)/2

Hence it is proved!!✔✔