The acceleration of a body moving with initial velocity u changes with distance x as a =k2 √x where K is a positive constant. The distance travelled by the body when its velocity becomes 2u is

a) (3u/2k)3/4

b) (3u/2k)4/3

c) (3u/2k)3/2

d) (3u/2k)2/3

Answer: (3u/2k)4/3

Solution:

Given, a =k2 √x 

vdv/dx = k2 √x 

⇒ vdv = k2 √x dx

\(\begin{array}{l}\int_{u}^{2u}vdv= \int_{x=0}^{x=x}k^{4}\sqrt{x}dx\end{array} \)

\(\begin{array}{l}\frac{v^{2}}{2}|_{u}^{2u} = (2/3)k^{4}x^{3/2}|_{0}^{x}\end{array} \)

(1/2) [(2u)2 – u2] = (2/3)k4x3/2

3u2/2 = (2/3)k4x3/2

x = (3u/2k)4/3

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