Question

Prove that, distance travelled = average velocity*time

Answer 1

Distance travelled(S)=average velocity{a=(v-u)/t}*time

From the definition of acceleration, we get, Velocity(v)=u(final velocity)+at(acceleretion*time)

∴ {(initial velocity+final velocity)/2}*time

∴ S ={(u+v)/2}*t &

v=u+at

∴ t=(v-u)/a

⇒S={(u+v/2}*{(v-u)/a}

    =(v²-u²)/2a

∴ v²-u² = 2aS

⇒ v²= 2aS+u²

⇒v²=u²+2as

Hence, by the above calculation it is proved that, distance travelled=average velocity*time

It should be noted here that,

Distance = S

Initial velocity = u

Final velocity = v

Time = t

Average velocity = a =(v-u)/t

Answer 2

Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2]  × [(v-u)/a]

or,  s = (v² – u²)/2a

or, 2as = v² – u²

or, v² = u² + 2as