Kinematics Question 372
Question: A particle starting with certain initial velocity and uniform acceleration covers a distance of 12 m in first 3 seconds and a distance of 30 m in next 3 seconds. The initial velocity of the particle is
Options:
A) $ 3m{{s}^{-1}} $
B) $ 2.5m{{s}^{-1}} $
C) $ 2m{{s}^{-1}} $
D) $ \text{1 }m{{s}^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let u be the initial velocity that have to find and a be the uniform acceleration of the particle. For $ t=3s $ , distance travelled
$ S=12m $ and For $ t=3+3=6s $
distance travelled $ S=12+30=42m $ From, $ S=ut+1/2at^{2} $
$ 12=u\times 3+\frac{1}{2}\times a\times 3^{2} $
Or $ 24=6y+9a~~~~~~~~~~~~~~~…( i ) $
Similarly, $ 42=u\times 6+\frac{1}{2}\times a\times 6^{2} $
Or $ 42=6y+18a~~~~~~~~~~~~~~~~~~…( ii ) $ On solving, we get $ u=1m{{s}^{-1}} $
BETA
