Question

A particle moves in a circle of radius 4 cm clockwise at constant speed 2 cm/s. If $\hat{x}\ and\ \hat{y}$ are unit acceleration vectors along X and Y-axis respectively (in cm/s²), the acceleration of the particle at the instant half way between P and Q is given by
(a) $-4(\hat{x}+\hat{y})$(b) $4(\hat{x}+\hat{y})$(c) ${-(\hat{x}+\hat{y})}/{\sqrt{2}}$(d) ${(\hat{x}-\hat{y})}/{4}$

Answer 1

Welcome student,

Corresponding image is missing in your question. I'll add that in attachment.

◆ Answer -

a = -(x+y)/√2 cm/s^2

● Explanation -

Magnitude of acceleration at mid P-Q is given by -

|a| = v^2 / r

|a| = 2^2 / 4

|a| = 1 cm/s^2

Particle will make an angle of 45° with X-axis & Y-axis when its midway P-Q.

Horizontal component of acceleration of the the particle is given by -

ax = -|a|.cosθ x

ax = -1 × cos45 x

ax = -x/√2

Vertical component of acceleration of the the particle is given by -

ay = -|a|.sinθ y

ay = -1 × sin45 y

ay = -y/√2

Total acceleration of the particle is given by -

a = ax + ay

a = (-1/√2 x) + (-1/√2 y)

a = -(x+y)/√2 cm/s^2

Thus, the acceleration of the particle at the instant half way between P and Q is -(x+y)/√2 cm/s^2.

Thanks for asking..