Applications Of Derivatives Question 144
Question: A particle moves along a straight line so that its distance s in time t sec is $ s=t+6t^{2}-t^{3} $ . After what time is the acceleration zero
[AMU 1999]
Options:
A) 2 sec
B) 3 sec
C) 4 sec
D) 6 sec
Show Answer
Answer:
Correct Answer: A
Solution:
$ s=t+6t^{2}-t^{3} $
therefore $ \frac{ds}{dt}=1+12t-3t^{2} $
therefore $ v=5,cm/\sec $
therefore $ \frac{d^{2}s}{dt^{2}}=0 $
therefore $ 12-6t=0\Rightarrow t=2 $ .
BETA
