Question

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Answer 1

Q : A particle moves rectilinearly with deceleration whose modulus depends on the velocity of v of a particle as a=k√v, where k is a positive constant. At the initial moment, the velocity of the point is v(o) . The distance travelled by the particle is :


Final Answer : (a)
$\frac{2 {v}^{ \frac{3}{2} } }{3k}$

Steps and Understanding:

1) Given motion is : Rectilinear Deceleration
We will write
Since, Below RHS is positive so, muliptly by (-1) to dv (which is negative) to make both sides positive.

$a = \frac{v (- dv)}{dx} = k \sqrt{v} \\ = > \frac{ - vdv}{dx} = k \sqrt{v} \\ = > - \sqrt{v} dv = kdx$

2) Then, we will integrate both sides.
Integrate LHS from v = v' to v = 0
RHS from x = 0 to x = x.

3) After Integrating,
we get
$x = \frac{2 {v}^{ \frac{3}{2} } }{3k}$

So, Distance travelled by the particle before it stops is
$\frac{2 {v}^{ \frac{3}{2} } }{3k}$

Answer 2

Explanation:

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