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) 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 ((v_1 + v_2 )/2) (3) Distance travelled in successive equal time intervals are in proportion of 1 : 3 : 5 ... 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 ((v_1 − v_2 )/(v_2 − v_3 )) = ((t_1 + t_2 )/(t_2 + t


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

$A particle starts from rest at t = 0 and$ $moves with uniform acceleration . Then$ $(1) In any time interval starting from$ $t = 0 the space - average of the velocity$ $is \frac{4}{3} times of time average velocity$ $(2) If v = v_{1} at t = t_{1} and v = v_{2} at t = t_{2}$ $then time average velocity between t_{1}$ $and t_{2} is \frac{v_{1} + v_{2}}{2}$ $(3) Distance travelled in successive$ $equal time intervals are in proportion$ $of 1 : 3 : 5 . . . and so on$ $(4) If v_{1}, v_{2}, v_{3} denote the average$ $velocities in three successive intervals$ $of time t_{1}, t_{2}, t_{3} then \frac{v_{1} - v_{2}}{v_{2} - v_{3}} = \frac{t_{1} + t_{2}}{t_{2} + t_{3}}$
	Answered by ajfour last updated on 25/Nov/17

	$(1) . {(v)}_{s p a c e a v g} = \frac{\int v d x}{Δ x}$ $a s \frac{v d v}{d x} = a \Rightarrow v d x = \frac{v^{2} d v}{a}$ $s o {(v)}_{s p a c e a v g} = \frac{\int_{0}^{v} v^{2} d v}{a Δ x}$ $= \frac{v^{3}}{3 a (v t / 2)} = \frac{2 v^{2}}{3 a t} = \frac{(4 a t)}{3 a t} (\frac{v}{2})$ ${(v)}_{s p a c e a v g} = \frac{4}{3} {(v)}_{t i m e a v g} .$ $(2) . {(v)}_{t i m e a v g} = \frac{s}{△ t}$ $= \frac{v_{1} △ t + \frac{1}{2} a {(△ t)}^{2}}{△ t}$ $= v_{1} + \frac{1}{2} (a △ t)$ $= \frac{2 v_{1} + (v_{2} - v_{1})}{2} = \frac{v_{1} + v_{2}}{2} .$ $(3) . s_{1} = \frac{1}{2} {a t}^{2}$ $s_{1} + s_{2} = \frac{1}{2} a {(2 t)}^{2} = 4 s_{1}$ $\Rightarrow s_{2} = 3 s_{1}$ $s_{1} + s_{2} + s_{3} = \frac{1}{2} a {(3 t)}^{2} = 9 s_{1}$ $\Rightarrow s_{3} = 5 s_{1} a n d s o o n . .$ $(4) . l e t v e l o c i t y a t t = t_{i} b e u$ $v_{1} = u + \frac{{a t}_{1}}{2}$ $v_{2} = u + {a t}_{1} + \frac{{a t}_{2}}{2}$ $v_{1} - v_{2} = - \frac{a}{2} (t_{1} + t_{2}) . . . . . (i)$ $v_{3} = u + {a t}_{1} + {a t}_{2} + \frac{1}{2} {a t}_{3}$ $s o v_{2} - v_{3} = - \frac{a}{2} (t_{2} + t_{3}) . . . . (i i)$ $f r o m (i) a n d (i i) w e g e t$ $\frac{v_{1} - v_{2}}{v_{2} - v_{3}} = \frac{t_{1} + t_{2}}{t_{2} + t_{3}} .$
	Commented by Tinkutara last updated on 25/Nov/17

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