Question

A particle moves in the xy-plane with coordinates given by

x = A cosωt and y = A sin ωt,

where A = 1. 5 m and ω = 2. 0 rad/s.

What is the magnitude of the particle's acceleration?

a. Zero

b. 1. 3 m/s2

c. 3. 0 m/s2

d. 4. 5 m/s2

e. 6. 0 m/s2

Answer 1

The position of the particle in the xy-plane is given by:

x = A cos(ωt)

y = A sin(ωt)

Differentiating the position equation twice with respect to time gives the acceleration equation:

a = -Aω^2cos(ωt) - Aω^2sin(ωt)

The magnitude of the acceleration is:

|a| = sqrt[(Aω^2cos(ωt))^2 + (Aω^2sin(ωt))^2]

Since the maximum value of sin and cos functions is 1, the maximum magnitude of acceleration is:

|a|max = sqrt[(Aω^2)^2 + (Aω^2)^2] = sqrt[2(Aω^2)^2] = sqrt[2]Aω^2

Substituting the given values gives:

|a|max = sqrt[2](1.5 m)(2.0 rad/s)^2 = 6.0 m/s^2

Therefore, the magnitude of the particle's acceleration is 6.0 m/s2, option (e).

Step-by-step explanation: