Question

A particle moves along a path y=2cos 3x in such a way that the component of it velocity in x direction is 4 units. find its acceleration

Answer 1

Given path,

$$y=2cos 3x$$

Differentiate with respect to x, We get

$$\frac {dy}{dx} = 2 \times ( - sin 3x ) \times 3$$

$$\frac {dy}{dx} = -6 \times sin 3x$$

But, We have

$$\frac {dy}{dx} = 4$$

So,

$$-6 \times sin 3x = 4$$

$$sin 3x = - \frac {2}{3}$$

We know ,

$$sin^2 3x + cos^2 3x = 1$$

$$cos^2 3x = 1 - sin^2 3x$$

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

$$cos 3x = \frac {\sqrt 5}{3}$$

$$Acceleration =\frac {d^2 y} {dx^2}$$

$$= \frac {d}{dx} ( - 6 \times sin 3x)$$

$$- 6 \times cos 3x \times 3 = - 18 \times cos 3x = - 18 \times \frac {\sqrt 5}{3} = - 6 \times {\sqrt 5}$$

Answer:

Acceleration = $- 6 \times \sqrt 5$