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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-




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.
Aparticlestartsfromrestattimet=0andmovesinastraightlinewithvariableaccelerationam/s2wherea=t5,0t5,a=t5+10t2,t5,tbeingmeasuredinseconds.Showthatthevelocityis2212m/swhent=5and11m/swhent=10.Showalsothatthedistancetravelledbytheparticleinthefirst10secondsis(431310ln2)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=0tt5dt;0t5=52+5t(t5+10t2)dt;t>5=52+(t21010t)(522)Att=5,v=212m/s=52m/s,andatt=10,v=11m/s.Distancesuntilt=10iss1+s2,s1=05(t210)dt=12530=256s2=510(t21010t+2)dt=(t33010lnt+2t)510=(100325610ln2+10)s1+s2=431310ln2.(Indeed!)
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)
x.=xdt..=05t5dt=2510.ms=2.5msAfter10secondsx.=2.5+510t5+10t2dt=7.5+1+2.5ms=11ms

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