Q.
A particle moves with constant acceleration along a straight line. If v1 , v2 and v3 are the average velocities in the three successive intervals t1 , t2 and t3 of time, then the correct relation is
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NTA AbhyasNTA Abhyas 2020Motion in a Straight Line
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Solution:
If v1′v2′ and v3′ be the velocities at the ends of time intervals t1,t1+t2 and t1+t2+t3 respectively, then v1′=u+at1,v2′=u+a(t1+t2),v3′=u+a(t1+t2+t3) v1=2u+v1′=2u+u+at1=u+21at1 v2=2v1′+v2′=2(u+at1)+[u+a(t1+t2)] =u+at1+21at2 v3=2u2′+v3′=2u+at1+at2+u+a(t1+t2+t3)
Or v3=u+at1+at2+21at3 v1−v2=u+21at1−u−at1−21at2=−21a(t1+t2) v2−v3=u+at1+21at2−u−at1−at2−21at3 =−21at2−21at3=−21a(t2+t3) v2−v3v1−v2=t2+t3t1+t2