Answer to Question #346683 in Physics for blueming

Question

The velocity of a particle undergoing simple harmonic motion is described by,

𝑣(𝑡) = 2.25m/s cos(0.555 rad/s∙ 𝑡 + 0.450 rad) What is the particle’s maximum acceleration? 

Answer 1

$$𝑣(𝑡) = 2.25 {\:\rm m/s} \cos\left((0.555 {\:\rm rad/s}) 𝑡 + 0.450 {\:\rm rad}\right)$$

The acceleration

$$a(t)=𝑣'(𝑡) = 2.25 {\:\rm m/s} *(0.555 {\:\rm rad/s}) \\ *\sin\left((0.555 {\:\rm rad/s}) 𝑡 + 0.450 {\:\rm rad}\right)$$

$$a_{\max}= 2.25 {\:\rm m/s} *(0.555 {\:\rm rad/s})=1.25\:\rm m/s^2$$