SATHEE: Kinematics Question 368

Kinematics Question 368

Question: An object, moving with a speed of $6.25 m / s$ , it’s decelerate data rate given by: $\frac{d v}{d t} = -2 .5 \sqrt{v}$ 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:

$\frac{d v}{d t} = - 2.5 \sqrt{v} \Rightarrow \frac{d v}{\sqrt{v}} = 2.5 d t$

$Integrating, \int lim i t^{'} s_{6.25}^{0} v^{\frac{- 1}{2}} d v = - 2.5 \int lim i t^{'} s_{0}^{t} d t$
$\Rightarrow [\frac{v^{+ \frac{1}{2}}}{(^{1} /_{2})}]_{6.25}^{0} = - 2.5 [t]_{0}^{t}$
$\Rightarrow - 2 {(6.25)}^{^{1} /_{2}} = - 2.5 t \Rightarrow t = 2 sec$

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Kinematics Question 368

Question: An object, moving with a speed of 6.25m/s , it’s decelerate data rate given by: dvdt= -2.5v where v it’s the instantaneous speed. The time taken by the object, to come to rest, would be

Options:

Answer:

Solution:

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Question: An object, moving with a speed of $6.25 m / s$ , it’s decelerate data rate given by: $\frac{d v}{d t} = -2 .5 \sqrt{v}$ where v it’s the instantaneous speed. The time taken by the object, to come to rest, would be