Question

A particle moves along a path y=2cos3x in such a way that the component of its velocity in the x-direction is 4units find its acceleration

Answer 1

The particle's acceleration is $-6\sqrt5$

Explanation:

Given:

The path of the particle is given as:

$y=2\cos 3x$

The 'x' component of velocity is, $\frac{dy}{dx}=4$ units

So, let us differentiate the above function with respect to 'x'. This gives,

$\frac{dy}{dx}=\frac{d}{dx}(2\cos 3x)\\\frac{dy}{dx}=2\frac{d}{dx}(\cos 3x)\\\frac{dy}{dx}=2(-\sin 3x)(3)\\\frac{dy}{dx}=-6\sin (3x)$

Now, $\frac{dy}{dx}=4$. So,

$-6\sin (3x)=4\\\sin (3x)=-\frac{4}{6}=-\frac{2}{3}$

Now, acceleration is given as the derivative of velocity with respect to 'x'. So,

Acceleration is given as:

$a=\frac{d}{dx}(\frac{dy}{dx})\\a=\frac{d}{dx}(-6\sin (3x))\\a=-6\cos (3x)(3)\\a=-18\cos (3x)$

Now, $\cos 3x$ can be found using trigonometric identity.

$sin^2(3x)+cos^2(3x)=1\\\cos(3x)=\sqrt{1-sin^2(3x)}\\\cos(3x)=\sqrt{1-(-\frac{2}{3})^2}\\\cos(3x)=\sqrt{1-\frac{4}{9}}\\\cos(3x)=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$

Therefore, the acceleration is given as:

$a=-18\cos (3x)=-18\times \frac{\sqrt5}{3}=-6\sqrt5$