Answer:
Part a)
\(a_r = 1.25 m/s^2\)
Part b)
\(a_t = 0.4 m/s^2\)
Part c)
\(a_{net} = 1.31 m/s^2\)
Explanation:
initial speed of the turn table at t = 0 is given as
\(v_i = 0\)
after t = 1.75 s the speed is given as
\(v_f = 0.700 m/s\)
now the tangential acceleration is given as
\(a_t = \frac{v_f - v_i}{t}\)
\(a_t = \frac{0.700 - 0}{1.75}\)
\(a_t = 0.4 m/s^2\)
Part a)
Now the speed of the disc after t = 1.25 s is given as
\(v_f = v_i + at\)
here we will have
\(v_f = 0 + (0.4)(1.25)\)
\(v_f = 0.5 m/s\)
now radial acceleration is given as
\(a_r = \frac{v^2}{R}\)
\(a_r = \frac{0.5^2}{0.20}\)
\(a_r = 1.25 m/s^2\)
Part b)
tangential acceleration is given as
\(a_t = 0.4 m/s^2\)
Part c)
total acceleration is given as
\(a_{net} = \sqrt{a_r^2 + a_t^2}\)
\(a_{net} = \sqrt{1.25^2 + 0.4^2}\)
\(a_{net} = 1.31 m/s^2\)
