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A-particle-starts-from-rest-at-time-t-0-and-moves-in-a-straightline-with-variable-acceleration-a-m-s-2-where-a-t-5-0-t-5-a-t-5-10-t-2-t-5-t-being-measured-in-seconds-

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Question Number 126092 by physicstutes last updated on 17/Dec/20
A particle starts from rest at time t = 0 and moves in a straightline with variable acceleration a m/s^2 where a = (t/5) , 0 ≤ t ≤ 5 , a = (t/5) + ((10)/t^2 ) , t ≥ 5, t being measured in seconds. Show that the velocity is 22(1/2) m/s when t = 5 and 11 m/s when t = 10. Show also that the distance travelled by the particle in the first 10 seconds is (43(1/3)−10 ln 2) m.
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{time}\:{t}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{moves}\:\mathrm{in}\: \\ $$$$\mathrm{a}\:\mathrm{straightline}\:\mathrm{with}\:\mathrm{variable}\:\mathrm{acceleration}\:{a}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{where}\: \\ $$$$\:{a}\:=\:\frac{{t}}{\mathrm{5}}\:,\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{5}\:,\:{a}\:=\:\frac{{t}}{\mathrm{5}}\:+\:\frac{\mathrm{10}}{{t}^{\mathrm{2}} }\:,\:{t}\:\geqslant\:\mathrm{5},\:{t}\:\mathrm{being}\:\mathrm{measured}\:\mathrm{in}\:\mathrm{seconds}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{is}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{m}/\mathrm{s}\:\mathrm{when}\:{t}\:=\:\mathrm{5}\:\mathrm{and} \\ $$$$\mathrm{11}\:\mathrm{m}/\mathrm{s}\:\mathrm{when}\:{t}\:=\:\mathrm{10}. \\ $$$$\mathrm{Show}\:\mathrm{also}\:\mathrm{that}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{10}\:\mathrm{seconds}\:\mathrm{is}\:\:\left(\mathrm{43}\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{10}\:\mathrm{ln}\:\mathrm{2}\right)\:\mathrm{m}. \\ $$
Answered by ajfour last updated on 17/Dec/20
v=∫_0 ^( t) (t/5)dt ; 0≤t≤5 = (5/2)+∫_5 ^( t) ((t/5)+((10)/t^2 ))dt ; t>5 =(5/2)+((t^2 /(10))−((10)/t))−((5/2)−2) ⇒ At t=5, v=2(1/2)m/s = (5/2)m/s, and at t=10, v= 11m/s. Distance s until t=10 is s_1 +s_2 , ∀ s_1 =∫_0 ^( 5) ((t^2 /(10)))dt=((125)/(30))=((25)/6) s_2 =∫_5 ^( 10) ((t^2 /(10))−((10)/t)+2)dt =((t^3 /(30))−10ln t+2t)∣_5 ^(10) =(((100)/3)−((25)/6)−10ln 2+10) s_1 +s_2 = 43(1/3)−10ln 2 . (Indeed!)
$${v}=\int_{\mathrm{0}} ^{\:\:{t}} \frac{{t}}{\mathrm{5}}{dt}\:\:\:;\:\:\:\:\mathrm{0}\leqslant{t}\leqslant\mathrm{5} \\ $$$$\:\:=\:\frac{\mathrm{5}}{\mathrm{2}}+\int_{\mathrm{5}} ^{\:{t}} \left(\frac{{t}}{\mathrm{5}}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }\right){dt}\:\:\:;\:\:{t}>\mathrm{5} \\ $$$$\:\:=\frac{\mathrm{5}}{\mathrm{2}}+\left(\frac{{t}^{\mathrm{2}} }{\mathrm{10}}−\frac{\mathrm{10}}{{t}}\right)−\left(\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{2}\right) \\ $$$$\:\Rightarrow\:\:{At}\:{t}=\mathrm{5},\:\:{v}=\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{m}/{s}\:=\:\frac{\mathrm{5}}{\mathrm{2}}{m}/{s}, \\ $$$${and}\:\:{at}\:\:{t}=\mathrm{10},\:\:\:{v}=\:\mathrm{11}{m}/{s}. \\ $$$${Distance}\:{s}\:{until}\:{t}=\mathrm{10}\:\:{is}\:{s}_{\mathrm{1}} +{s}_{\mathrm{2}} \:,\:\forall \\ $$$${s}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\:\mathrm{5}} \:\left(\frac{{t}^{\mathrm{2}} }{\mathrm{10}}\right){dt}=\frac{\mathrm{125}}{\mathrm{30}}=\frac{\mathrm{25}}{\mathrm{6}} \\ $$$$\:\:{s}_{\mathrm{2}} =\int_{\mathrm{5}} ^{\:\mathrm{10}} \left(\frac{{t}^{\mathrm{2}} }{\mathrm{10}}−\frac{\mathrm{10}}{{t}}+\mathrm{2}\right){dt} \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{t}^{\mathrm{3}} }{\mathrm{30}}−\mathrm{10ln}\:{t}+\mathrm{2}{t}\right)\mid_{\mathrm{5}} ^{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:=\left(\frac{\mathrm{100}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{6}}−\mathrm{10ln}\:\mathrm{2}+\mathrm{10}\right) \\ $$$${s}_{\mathrm{1}} +{s}_{\mathrm{2}} =\:\mathrm{43}\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{10ln}\:\mathrm{2}\:.\:\:\left({Indeed}!\right) \\ $$
Answered by Dwaipayan Shikari last updated on 17/Dec/20
x^. =∫x^(..) dt = ∫_0 ^5 (t/5)dt= ((25)/(10)).(m/s)=2.5(m/s) After 10 seconds x^. =2.5+∫_5 ^(10) (t/5)+((10)/t^2 )dt=7.5+1 +2.5 (m/s) =11(m/s)
$$\overset{.} {{x}}=\int\overset{..} {{x}dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{5}} \frac{{t}}{\mathrm{5}}{dt}=\:\frac{\mathrm{25}}{\mathrm{10}}.\frac{{m}}{{s}}=\mathrm{2}.\mathrm{5}\frac{{m}}{{s}} \\ $$$${After}\:\mathrm{10}\:{seconds}\: \\ $$$$\overset{.} {{x}}=\mathrm{2}.\mathrm{5}+\int_{\mathrm{5}} ^{\mathrm{10}} \frac{{t}}{\mathrm{5}}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }{dt}=\mathrm{7}.\mathrm{5}+\mathrm{1}\:+\mathrm{2}.\mathrm{5}\:\frac{{m}}{{s}}\:=\mathrm{11}\frac{{m}}{{s}} \\ $$

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