Respuesta :
A)The speed of the rider is 2.6m/s
B)The rider's radial acceleration is [tex]2.75m/s^2[/tex]
C)The radial acceleration when the time for one rotation is halved is [tex]10.7 m/s^2[/tex]
Explanation:
Time for one rotation t=6s
[tex]diamter\ of\ the\ ride=16\ feet=16\times 0.3048=4.9 m\\radius\ r=diameter/2=4.9/2=2.45 m[/tex]
speed is the rate of change of distance with time.
to calculate the distance we have to find the circumference of the ride
A)[tex]circumference=2\pi r=2\times3.14\times 2.45\\=15.4\\speed\ of\ the\ ride\ v=circumference/time=15.4/6=2.6m/s[/tex]
B)
[tex]radial\ acceleration=v^2/r\\=2.6^2/2.45=2.75 m/s^2[/tex]
C)If the time of rotation is halved, speed/velocity as well as the radial acceleration changes.
new time t'=3s
new velocity v'=circumference/t'=15.4/3=5.13 m/s[tex]new\ radial\ acceleration=v'^2/r=5.13^2/2.45=10.7m/s^2[/tex]
Merry go round is the ride for kids and teenagers to enjoy the time.
According to the question, the speed of the rider are as follows:-
The time required for one rotation is 6seconds, the radius of merry go round is 2.45m/s because the diameter of the merry go round is:-
[tex]16 *0.3048 = 4.9m[/tex]
To calculate the speed, we have to find the circumference of the merry go round,
Then, [tex]2\pi r= 2*3.14*2.45 =15.4[/tex] m
Hence, speed is [tex]\frac{15.4}{6} = 2.6m/s[/tex]
The change in velocity is defined as the accerlation. The formula for the radial acceleration is :-[tex]\frac{v^2}{r}[/tex]
[tex]\frac{2.6^2}{2.45} =2.75m/s[/tex]
The acceleration for the new velocity is:-
[tex]\frac{15.4}{3} =5.13m/s[/tex]
[tex]hence,\ acceleration \ is \frac{v^2}{r}[/tex]
[tex]\frac{5.13^2}{2.45}[/tex] =10.7m/s
These are the answer to the question.
For more information, refer to the link:-
https://brainly.com/question/19359821
