Applications Of Derivatives Question 67
Question: A particle moves so that $ S=6+48t-t^{3} $ . The direction of motion reverses after moving a distance of
[Kurukshetra CEE 1998]
Options:
A) 63
B) 104
C) 134
D) 288
Show Answer
Answer:
Correct Answer: C
Solution:
$ S=6+48t-t^{3} $
$ v=\frac{dS}{dt}=0+48-3t^{2} $ when direction of motion reverses, $ v=0 $
$ 48-3t^{2}=0 $
$ \Rightarrow $ $ t=-4,\ 4 $ ; $ {{(S)}_4}=6+192-64 $ i.e., $ S=134 $ .
BETA
