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THE DISPLACEMENT OF A PARTICLE STARTING from rest and moving under a constant acceleration is calculated by the formula S= 1/2 at^2 .if there occurs an error of 30% in the measurement of time interval , what error will be introduced in the calculation of S?
40.82%
i think i answered this question didnt u get it..?
here S= 1/2at² but due to error put S+dS instead of S and t+dt instead of t (here d is delta) hence... S+dS=1/2a(t+dt)² =1/2a(t²+2tdt+[dt]²) .... here S=1/2at² hence they get cancelled out... we get here is dS=atdt+1/2a[dt]².....divide it by S=1/2at²...u would get dS/S=2dt/t+[dt/t]²...(1) it is given that dt/t=30/100=0.3 ... now put this in equation (1).. u would get dS/S=0.6+0.09=0.69=69% ... and that is the answer..
Given that (S= 1/2 at^2) and error is 30% S`=(1/2) a(30*t/100)^2 =(1/2)*a*0.09*t^2 S`=0.09S % error in S=(S-S`/S)*100 =((1-0.09)S/S)*100 =91%
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