Question

A particle is performing motion whose position vector is
given by r = 10 (cos pt i + sin ptſ)m, when p is in
rad/s and t is in seconds. The magnitude of
centripetal acceleration at t = 4 s, is​

Answer 1

Answer:

10p² m/s

Explanation:

Given the position vector of the particle

r = 10 (cos(pt) i + sin(pt) j)

Differentiating r w.r.t. t we get the velocity vector as

v = 10p(-sin(pt) i + cos(pt) j)

Differentiating again w.r.t. t we get the acceleration vector as

a = 10p²(-cos(pt) i - sin(pt) j)

The magnitude of acceleration

= $\sqrt{[10p^{2} (cos^{2}pt+sin^{2}pt)]^{2}}$

= 10p²

Answer 2

Given the position vector of the particle

r = 10 (cos(pt) i + sin(pt) j)

Differentiating r w.r.t. t we get the velocity vector as

v = 10p(-sin(pt) i + cos(pt) j)

Differentiating again w.r.t. t we get the acceleration vector as

a = 10p²(-cos(pt) i - sin(pt) j)

The magnitude of acceleration  = 10p2