Answer:
\(v = 0.24 m/s\)
\(a = 6.75 \times 10^{-4} m/s^2\)
Explanation:
Given that wheel completes one round in total time T = 37.3 min
so angular speed of the wheel is given as
\(\omega = \frac{2\pi}{T}\)
\(\omega = \frac{2\pi}{37.3} rad/min\)
now the angle turned by the wheel in time interval of t = 7.40 min
\(\theta = \omega t\)
\(\theta = (\frac{2\pi}{37.3})(7.40) = 0.4\pi\)
PART 1)
Now the average velocity is defined as the ratio of displacement and time
here displacement in given time interval is
\(d = 2Rsin\frac{\theta}{2}\)
R = radius = 91.5 m
\(d = 183sin(0.2\pi) = 106.8 m\)
Now time to turn the wheel is given as
\(t = 7.40 min = 444 s\)
now we have
\(v = \frac{d}{t} = \frac{106.8}{444}\)
\(v = 0.24 m/s\)
PART 2)
Now average acceleration is defined as ratio of change in velocity in given time interval
here velocity of a point on its rim is given as
\(v = R\omega\)
\(v = (91.5)(\frac{2\pi}{37.3\times 60})\)
\(v = 0.257 m/s\)
now change in velocity when wheel turned by the above mentioned angle is given as
\(\Delta v = 2vsin\frac{\theta}{2}\)
\(\Delta v = 2(0.257)sin(0.2\pi)\)
\(\Delta v = 0.3 m/s\)
time interval is given as
\(t = 7.40 min = 444 s\)
now average acceleration is given as
\(a = \frac{0.3}{444}\)
\(a = 6.75 \times 10^{-4} m/s^2\)
