CBSE Class 11-science Answered
A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s.afterwards it decelrates with the constant rate and comes to rest .If the total time taken is 4 S, the distance travlled is
Asked by simashreegogoi014 | 25 Jul, 2019, 05:10: PM
Expert Answer
It is assumed magnitude of acceleration and that of deceleration are same.
In that case particle is at 2 s in accelerated motion and for 2 second in decelerated motion.
Staring from rest, if particle attains speed 8 m/s, then acceleration is given by, v = a×t or 8 = a×2 or a =4 m/s2
hence distance S1 travelled with acceleration, S1 = (1/2)×a×t2 = (1/2)×4×22 = 8 m
distance S2 travelled with deceleration, S2 = u×t - (1/2)×a×t2 = 8×2 - (1/2)×4×22 = 8 m
total distance, S1 + S2 = 8 + 8 = 16m
Answered by Thiyagarajan K | 25 Jul, 2019, 06:29: PM
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