Question

An object moves in a stright line with deceleration whose magnitude varies with velocity as 3vpower2/-3if at an initial point the velocity is 8m/sec then the distance travelled by the object before it stops

Answer 1

Correct question :-

A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as w = α√v, where α is a positive constant. At the initial moment the velocity of the point is equal to vo. What distance will it traverse before it stops? What time will it take to cover that distance?

Solution :-

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.

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

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:

$\displaystyle{x = \frac{2 {v}^{ \frac{3}{2} } }{3k} }$

So, Distance travelled by the particle before it stops is:

$\boxed{\displaystyle{ \frac{2 {v}^{ \frac{3}{2} } }{3k} }}$