Applications Of Derivatives Question 417
Question: The motion of a particle is described as $ s=2-3t+4t^{3} $ . What is the acceleration of the particle at the point where its velocity is zero-
Options:
A) 0
B) 4 unit
C) 8 unit
D) 12 unit
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given rule is: Distance, $ s=2-3t+4t^{3} $
$ \Rightarrow $ Velocity $ =\frac{ds}{dt}=-3+12t^{2} $
$ \Rightarrow $ Acceleration $ =\frac{d^{2}s}{dt^{2}}=16t $ Since, velocity is zero
$ \therefore \frac{ds}{dt}=0;,\Rightarrow 0=-3+12t^{2} $
$ \Rightarrow t=\frac{\sqrt{3}}{12}=\frac{1}{2}\Rightarrow \frac{d^{2}s}{dt^{2}}=16t $ Acceleration (when velocity is zero) $ =16\times \frac{1}{2}=8 $ unit.
BETA
