Question

9.
A particle is performing motion whose position vector is
given by r = 10 (cospt i + sin pt j)m, when p is in
rad/s and t is in seconds. The magnitude of
centripetal acceleration at t = 4 s, is
(1) 30p2 m/s2 (2) 10p2 m/s2
(3) 45p2 m/s2 (4) 90p2 m/s2
​

Answer 1

Answer ⇒ Option (2). is correct.

Explanation ⇒ Given Position Vectors are,

R = 10(Cospt i + Sin pt j)  [Let r = R.]

In x direction,

Rx = 10cospt

Differentiating both sides with respect to t,

Vx= 10(-Sinpt) × p

Vx = -10pSinpt.

Again Differentiating both sides with respect to t,

ax = -10p²Cospt.

For y direction,

Ry = 10Sinpt

Differentiating both sides with respect to t,

Vy = 10pCospt

Again, Differentiating both sides with respect to t,

ay = -10p²Sinpt.

Net acceleration = √(ax² + ay²)

Net acceleration = √(100p⁴Cos²pt + 100p⁴Sin²pt)

Net acceleration = 10p²√(Sin²pt + Cos²pt)

∴ Net acceleration = 10p²√(1)

∴ Net acceleration = 10p²

Hence, Option (2). is correct.