An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by where v is the instantaneous speed. The time taken by the object, to come to rest, would be :

Engineering

Physics

Straight Line Motion

Question

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by

$\frac{dv}{dt} = - 2.5 \sqrt{v}$

where v is the instantaneous speed. The time taken by the object, to come to rest, would be :

1 s

2 s

4 s

8 s

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$\int\limits_{6.25}^0 {\frac{{dv}}{{\sqrt v }} = - 2.5\int\limits_0^t {dt} } $

$\left| {2\sqrt \upsilon } \right|_{6.25}^0 = - 2.5t$

$2\sqrt {6.25} = 2.5t$

t = 2 sec.

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