Answer:
The the magnitude of the maximum velocity and acceleration are 16.93 m/s and [tex]5.213\times10^{3}\ m/s^2[/tex].
Explanation:
Given that,
Rate = 2940 rev/min
Amplitude = ±5.50 cm
(a). We need to calculate the magnitude of the maximum velocity
Using formula of maximum velocity
[tex]v=A\times\omega[/tex]
Where, A = amplitude
Put the value into the formula
[tex]v=5.50\times10^{-2}\times2940\times\dfrac{2\pi}{60}[/tex]
[tex]v=16.93\ m/s[/tex]
(b). We need to calculate the maximum acceleration
Using formula of maximum acceleration
[tex]a=A\times\omega^2[/tex]
[tex]a=5.50\times10^{-2}\times(2940\times\dfrac{2\pi}{60})^2[/tex]
[tex]a=5.213\times10^{3}\ m/s^2[/tex]
Hence, The the magnitude of the maximum velocity and acceleration are 16.93 m/s and [tex]5.213\times10^{3}\ m/s^2[/tex].
