The equation of a tangent to the parabola `y^2=""8x""i s""y""=""x""+""2` . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is (1) `(-1,""1)` (2) `(0,""2)` (3) `(2,""4)` (4) `(-2,""0)`

Options

A. The force at t = (1/8) s on the particle is `-16 pi^(2) A-m`
B. The particle is acted upon by on impulse of magnitude `4pi^(2)A-m` at t = 0 s and t (1/4) s
C. The particle is not acted upon by any force
D. The particle is not acted upon by a constant force

Answer

A, B, D

Explanation

x = 0 for t `lt `0 s.
x(t) = A sin `pit` , for `0 lt t lt (1)/(4)`s
x = 0 , for t `gt (1)/(4) s`
For , `0 lt t lt (1)/(4)s " " v(t) = (dx)/(dt) = 4 pi A cos 4 pi t `
a(t) = acceleration
= `(dv(t))/(dt) = -16 pi^(2) A sin 4 pi t `
At t = `(1)/(8) s , a(t) = - 16 pi^(2) A sin 4 pi xx (1)/(8) = - 16 pi^(2) A `
F = ma(t) = `-16 pi^(2) A xx m = - 16 pi^(2)mA`
Impulse = Change in linear momentum
=` F xx t = (-16 pi^(2) Am) xx (1)/(4)`
= `4 pi^(2)Am`
The impulse (Change in linear momentum)
at t = 0 is same as , t = `(1)/(4)` s .
Clearly , force depends upon A which is not constant . Hence , force is also not constant .