Kinematics Question 368
Question: An object, moving with a speed of $ 6.25m/s $ , it’s decelerate data rate given by: $ \frac{dv}{dt}=\text{ -2}\text{.5}\sqrt{v} $ where v it’s the instantaneous speed. The time taken by the object, to come to rest, would be
Options:
A) 2s
B) 4s
C) 8s
D) 1s
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dv}{dt}=-2.5\sqrt{v}\Rightarrow \frac{dv}{\sqrt{v}}=2.5dt $
$ \text{Integrating, }\int\lim it’s _{6.25}^{0}{{{v}^{\frac{-1}{2}}}}dv=-2.5\int\lim it’s _{0}^{t}{dt} $
$ \Rightarrow [ \frac{{{v}^{+\frac{1}{2}}}}{( {}^{1}/{} _{2} )} ] _{6.25}^{0}=-2.5[ t ] _{0}^{t} $
$ \Rightarrow -2{{( 6.25 )}^{{}^{1}/{} _{2}}}=-2.5t\Rightarrow t=2\text{ sec} $
