Applications Of Derivatives Question 163
Question: The equation of motion of a particle is given by $ s=2t^{3}-9t^{2}+12t+1 $ ,where s and t are measured in cm and sec. The time when the particle stops momentarily is
Options:
A) 1 sec
B) 2 sec
C) 1, 2 sec
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{ds}{dt}=6t^{2}-18t+12 $ = velocity = 0 (when particle stopped)
therefore $ 6t^{2}-18t+12=0\Rightarrow (t-1),(t-2)=0 $
Hence time 1, 2 sec.
BETA
