Question

a partical moves with a constant speed u along the curve y=sin x. the magnitude of acceleration at the point corresponding to x=π/2 is..

Answer 1

Answer:

At point π/2 the acceleration of particle is -1m/s².

Explanation:

Given : curve y=sin xvelocity is given by

$u=\frac{dy}{dx}\\\\u= \frac{d(sin x)}{dx} \\\\u= cos x$

Now ,we know acceleration is given as  

$a = \frac{du}{dx}a=\frac{d(cos x)}{dx} \\\\a= -sin x$

at point x=π/2

$a=-sin \frac{\pi }{2} \\\\a= -1$

Hence acceleration at point π/2 is a=  -1 m/s²