Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Find the below attachment
Actual displacement that he required to move
towards North
Displacement that he moved due to snow is
at 47 degree North of East
now in vector component form we can say
now the displacement that is more required to reach the destination is given as
so the magnitude of the displacement is given as
its direction is given as
so it is 5.54 km towards 5.38 degree North of West.