Question

a particle starts from origin and moves along a path whose equation is y=2x^(3/2) the distance travelled by particle when it reaches at x=7m is 1022/(3n). find the value of n.

Answer 1

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Answer 2

A particle starts from origin and moves along a path whose equation is y = 2x³/² the distance travelled by particle when it reaches at x = 7m is 1022/3n .

We have to find the value of n.

• The particle is moving along the equation y = 2x³/². so the distance travelled by particle will be the distance of particle from final position to initial position on y = 2x³/².

at x = 7 m , y = 2(7)³/²

so final position {7, 2(7)³/²} and initial position (0,0) [ because it starts from origin]

using distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$x_2=7,x_1=0,y_2=2(7)^{3/2},y=0$

$\implies d=\sqrt{(7-0)^2+(2(7)^{3/2}-0)^2}$

$\implies d=\sqrt{49+4\times343}$

$\implies d=37.696$

but the distance travelled by particle is given 1022/3n

so, 37.696 = 1022/3n

⇒n = 1022/(3 × 37.696) ≈ 9

Therefore the value of n is 9.