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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 (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_3 ))

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{moves}\:\mathrm{with}\:\mathrm{uniform}\:\mathrm{acceleration}.\:\mathrm{Then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{In}\:\mathrm{any}\:\mathrm{time}\:\mathrm{interval}\:\mathrm{starting}\:\mathrm{from} \\ $$$${t}\:=\:\mathrm{0}\:\mathrm{the}\:\mathrm{space}-\mathrm{average}\:\mathrm{of}\:\mathrm{the}\:\mathrm{velocity} \\ $$$$\mathrm{is}\:\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{times}\:\mathrm{of}\:\mathrm{time}\:\mathrm{average}\:\mathrm{velocity} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{If}\:{v}\:=\:{v}_{\mathrm{1}} \:\mathrm{at}\:{t}\:=\:{t}_{\mathrm{1}} \:\mathrm{and}\:{v}\:=\:{v}_{\mathrm{2}} \:\mathrm{at}\:{t}\:=\:{t}_{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{time}\:\mathrm{average}\:\mathrm{velocity}\:\mathrm{between}\:{t}_{\mathrm{1}} \\ $$$$\mathrm{and}\:{t}_{\mathrm{2}} \:\mathrm{is}\:\frac{{v}_{\mathrm{1}} \:+\:{v}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Distance}\:\mathrm{travelled}\:\mathrm{in}\:\mathrm{successive} \\ $$$$\mathrm{equal}\:\mathrm{time}\:\mathrm{intervals}\:\mathrm{are}\:\mathrm{in}\:\mathrm{proportion} \\ $$$$\mathrm{of}\:\mathrm{1}\::\:\mathrm{3}\::\:\mathrm{5}\:...\:\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{If}\:{v}_{\mathrm{1}} ,\:{v}_{\mathrm{2}} ,\:{v}_{\mathrm{3}} \:\mathrm{denote}\:\mathrm{the}\:\mathrm{average} \\ $$$$\mathrm{velocities}\:\mathrm{in}\:\mathrm{three}\:\mathrm{successive}\:\mathrm{intervals} \\ $$$$\mathrm{of}\:\mathrm{time}\:{t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,\:{t}_{\mathrm{3}} \:\mathrm{then}\:\frac{{v}_{\mathrm{1}} \:−\:{v}_{\mathrm{2}} }{{v}_{\mathrm{2}} \:−\:{v}_{\mathrm{3}} }\:=\:\frac{{t}_{\mathrm{1}} \:+\:{t}_{\mathrm{2}} }{{t}_{\mathrm{2}} \:+\:{t}_{\mathrm{3}} } \\ $$

Answered by ajfour last updated on 25/Nov/17

(1). (v)_(spaceavg) =((∫vdx)/(Δx)) as ((vdv)/dx)=a ⇒ vdx = ((v^2 dv)/a) so (v)_(spaceavg) =((∫_0 ^( v) v^2 dv)/(aΔx)) =(v^3 /(3a(vt/2)))=((2v^2 )/(3at))=(((4at))/(3at))((v/2)) (v)_(space avg) = (4/3)(v)_(time avg) . (2). (v)_(time avg) =(s/(△t)) =((v_1 △t+(1/2)a(△t)^2 )/(△t)) =v_1 +(1/2)(a△t) =((2v_1 +(v_2 −v_1 ))/2) =((v_1 +v_2 )/2) . (3). s_1 =(1/2)at^2 s_1 +s_2 =(1/2)a(2t)^2 =4s_1 ⇒ s_2 =3s_1 s_1 +s_2 +s_3 =(1/2)a(3t)^2 =9s_1 ⇒ s_3 =5s_1 and so on.. (4). let velocity at t=t_i be u v_1 =u+((at_1 )/2) v_2 =u+at_1 +((at_2 )/2) v_1 −v_2 =−(a/2)(t_1 +t_2 ) .....(i) v_3 =u+at_1 +at_2 +(1/2)at_3 so v_2 −v_3 =−(a/2)(t_2 +t_3 ) ....(ii) from (i) and (ii) we get ((v_1 −v_2 )/(v_2 −v_3 )) =((t_1 +t_2 )/(t_2 +t_3 )) .

$$\left(\mathrm{1}\right).\:\:\:\:\:\:\left({v}\right)_{{spaceavg}} =\frac{\int{vdx}}{\Delta{x}} \\ $$$${as}\:\:\:\frac{{vdv}}{{dx}}={a}\:\:\Rightarrow\:\:\:\:{vdx}\:=\:\frac{{v}^{\mathrm{2}} {dv}}{{a}} \\ $$$${so}\:\:\:\left({v}\right)_{{spaceavg}} =\frac{\int_{\mathrm{0}} ^{\:\:{v}} {v}^{\mathrm{2}} {dv}}{{a}\Delta{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{v}^{\mathrm{3}} }{\mathrm{3}{a}\left({vt}/\mathrm{2}\right)}=\frac{\mathrm{2}{v}^{\mathrm{2}} }{\mathrm{3}{at}}=\frac{\left(\mathrm{4}{at}\right)}{\mathrm{3}{at}}\left(\frac{{v}}{\mathrm{2}}\right) \\ $$$$\:\:\left({v}\right)_{{space}\:{avg}} =\:\frac{\mathrm{4}}{\mathrm{3}}\left({v}\right)_{{time}\:{avg}} \:. \\ $$$$\left(\mathrm{2}\right).\:\:\:\:\:\left({v}\right)_{{time}\:{avg}} \:=\frac{{s}}{\bigtriangleup{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{v}_{\mathrm{1}} \bigtriangleup{t}+\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\bigtriangleup{t}\right)^{\mathrm{2}} }{\bigtriangleup{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={v}_{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\left({a}\bigtriangleup{t}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{v}_{\mathrm{1}} +\left({v}_{\mathrm{2}} −{v}_{\mathrm{1}} \right)}{\mathrm{2}}\:=\frac{{v}_{\mathrm{1}} +{v}_{\mathrm{2}} }{\mathrm{2}}\:. \\ $$$$\left(\mathrm{3}\right).\:\:\:\:\:{s}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{s}_{\mathrm{1}} +{s}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\mathrm{2}{t}\right)^{\mathrm{2}} =\mathrm{4}{s}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\:\:\:{s}_{\mathrm{2}} \:=\mathrm{3}{s}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:{s}_{\mathrm{1}} +{s}_{\mathrm{2}} +{s}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\mathrm{3}{t}\right)^{\mathrm{2}} =\mathrm{9}{s}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:{s}_{\mathrm{3}} =\mathrm{5}{s}_{\mathrm{1}} \:\:\:\:\:\:\:{and}\:{so}\:{on}.. \\ $$$$\left(\mathrm{4}\right).\:\:\:{let}\:{velocity}\:{at}\:{t}={t}_{{i}} \:\:{be}\:{u} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{v}_{\mathrm{1}} ={u}+\frac{{at}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{v}_{\mathrm{2}} ={u}+{at}_{\mathrm{1}} +\frac{{at}_{\mathrm{2}} }{\mathrm{2}} \\ $$$${v}_{\mathrm{1}} −{v}_{\mathrm{2}} =−\frac{{a}}{\mathrm{2}}\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right)\:\:\:\:\:\:.....\left({i}\right) \\ $$$${v}_{\mathrm{3}} ={u}+{at}_{\mathrm{1}} +{at}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{at}_{\mathrm{3}} \\ $$$${so}\:\:\:{v}_{\mathrm{2}} −{v}_{\mathrm{3}} =−\frac{{a}}{\mathrm{2}}\left({t}_{\mathrm{2}} +{t}_{\mathrm{3}} \right)\:\:\:....\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$$\frac{{v}_{\mathrm{1}} −{v}_{\mathrm{2}} }{{v}_{\mathrm{2}} −{v}_{\mathrm{3}} }\:=\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{{t}_{\mathrm{2}} +{t}_{\mathrm{3}} }\:. \\ $$

Commented by Tinkutara last updated on 25/Nov/17

Thank you Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

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