An object, moving with a speed of $6.25\text{ m/s}$, 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 :

JEE Main · 2011 · 10 Apr 2011 · Offline · Kinematics

An object, moving with a speed of

6.25 m/s

, is decelerated at a rate given by :

\frac{d v}{d t} = - 2.5 v

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

(A)2 s✓ Correct
(B)4 s
(C)8 s
(D)1 s

Answer: (A)

Solution

Separating variables in

\frac{d v}{d t} = - 2.5 v

gives

\frac{d v}{v} = - 2.5 d t

, and integrating,

2 v = - 2.5 t + C

. At

t = 0

v = 6.25

, so

2 6.25 = 5 = C

. The body stops when

v = 0

, i.e.

0 = 5 - 2.5 t

, giving

t = 2 s

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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