A-particle-starts-from-rest-at-t-0-and-moves-with-uniform-acceleration-Then-1-In-any-time-interval-starting-from-t-0-the-space-average-of-the-velocity-is-4-3-times-of-time-average-velocity-2

Question Number 24687 by Tinkutara last updated on 24/Nov/17

A particle starts from rest at t = 0 and

moves with uniform acceleration . Then

(1) In any time interval starting from

t = 0 the space - average of the velocity

is \frac{4}{3} times of time average velocity

(2) If v = v_{1} at t = t_{1} and v = v_{2} at t = t_{2}

then time average velocity between t_{1}

and t_{2} is \frac{v_{1} + v_{2}}{2}

(3) Distance travelled in successive

equal time intervals are in proportion

of 1 : 3 : 5 \dots and so on

(4) If v_{1}, v_{2}, v_{3} denote the average

velocities in three successive intervals

of time t_{1}, t_{2}, t_{3} then \frac{v_{1} - v_{2}}{v_{2} - v_{3}} = \frac{t_{1} + t_{2}}{t_{2} + t_{3}}

Answered by ajfour last updated on 25/Nov/17

(1) . {(v)}_{s p a c e a v g} = \frac{\int v d x}{Δ x}

a s \frac{v d v}{d x} = a \Rightarrow v d x = \frac{v^{2} d v}{a}

s o {(v)}_{s p a c e a v g} = \frac{\int_{0}^{v} v^{2} d v}{a Δ x}

= \frac{v^{3}}{3 a (v t / 2)} = \frac{2 v^{2}}{3 a t} = \frac{(4 a t)}{3 a t} (\frac{v}{2})

{(v)}_{s p a c e a v g} = \frac{4}{3} {(v)}_{t i m e a v g} .

(2) . {(v)}_{t i m e a v g} = \frac{s}{△ t}

= \frac{v_{1} △ t + \frac{1}{2} a {(△ t)}^{2}}{△ t}

= v_{1} + \frac{1}{2} (a △ t)

= \frac{2 v_{1} + (v_{2} - v_{1})}{2} = \frac{v_{1} + v_{2}}{2} .

(3) . s_{1} = \frac{1}{2} {a t}^{2}

s_{1} + s_{2} = \frac{1}{2} a {(2 t)}^{2} = 4 s_{1}

\Rightarrow s_{2} = 3 s_{1}

s_{1} + s_{2} + s_{3} = \frac{1}{2} a {(3 t)}^{2} = 9 s_{1}

\Rightarrow s_{3} = 5 s_{1} a n d s o o n . .

(4) . l e t v e l o c i t y a t t = t_{i} b e u

v_{1} = u + \frac{{a t}_{1}}{2}

v_{2} = u + {a t}_{1} + \frac{{a t}_{2}}{2}

v_{1} - v_{2} = - \frac{a}{2} (t_{1} + t_{2}) \dots . . (i)

v_{3} = u + {a t}_{1} + {a t}_{2} + \frac{1}{2} {a t}_{3}

s o v_{2} - v_{3} = - \frac{a}{2} (t_{2} + t_{3}) \dots . (i i)

f r o m (i) a n d (i i) w e g e t

\frac{v_{1} - v_{2}}{v_{2} - v_{3}} = \frac{t_{1} + t_{2}}{t_{2} + t_{3}} .

Commented by Tinkutara last updated on 25/Nov/17

Thank you Sir!