Applications Of Derivatives Question 192
Question: The velocity of a particle at time t is given by the relation $ v=6t-\frac{t^{2}}{6} $ . The distance traveled in 3 seconds is, if $ s=0 $ at $ t=0 $
Options:
A) $ \frac{39}{2} $
B) $ \frac{57}{2} $
C) $ \frac{51}{2} $
D) $ \frac{33}{2} $
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Answer:
Correct Answer: C
Solution:
$ \frac{ds}{dt}=6t-\frac{t^{2}}{6} $ Now on integrating both sides $ s=3t^{2}-\frac{t^{2}}{18}+ $ constant , (where s is distance) Now put $ t=0 $ , then $ s=0 $ gives constant equal to 0 and putting $ t=3 $ , we get $ s=3{{(3)}^{2}}-\frac{3^{3}}{18}=27-\frac{27}{18}=\frac{51}{2} $ . Aliter : $ \int_0^{s}{ds}=\int_0^{3}{( 6t-\frac{t^{2}}{6} ),dt}=\frac{51}{2} $ .
BETA
