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If a particle starting from rest has an acceleration that increases linearly with time as a=2t, then the distance travelled in third second will be



Given that
now we know that 
 \frac{dv}{dt} =a
dv=2t dt
 \int\limits^v_0 {} \, dv= \int\limits^t_0 {2t} \, dt
v= t^{2}
now we know that 
 \frac{ds}{dt} =vds= t^{2} dt
 \int\limits^s_0 {} \, ds= \int\limits^t_0 { t^{2}}\, dt
s= \frac{ t^{3} }{3}
now in three seconds it will travel
 S_{3} = \frac{ 3^{3} }{3}
 S_{3} = \frac{27}{3}
in 2 seconds it will travel
 S_{2}= \frac{8}{3}
therefore in third second it will travel
 S_{3rd}= S_{3} -S_{12}
 S_{3rd} = \frac{19}{3} answer.............

1 5 1
ans isn't 15 its 19/3 but not getting how to solve
ok i will try
hey got it

 \frac{dv}{dt} =a=2t
integrating on both sides
c is the integration constant
The particle has started from rest 
so at velocity at t=0 is zero
integrating again omn both sides we displacement
x= \frac{t^3}{2}+C
where C is an integration constant
the distance travelled in 3rd second
==x(3)-x(2)= \frac{27}{3}- \frac{8}{3}= \frac{19}{3}
0 0 0
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