Question

What are:

a.) the angular speed
b.) the radial acceleration and
c.) the tangential acceleration of a spaceship negotiating
acircular turn of radius 3220 km at a speed of 29,000 km/h?

Answer 1

a) 0.0025 rad/s

b) 20.15 m/s2

c) 0 m/s2

Step-by-step explanation:

Unit conversion:

3220 km = 3220000m

29000km/h = 29000 km/h * 1000m/km * 1/3600h/s = 8056 m/s

a) The angular speed:

`\omega = (v)/(R) = (8056)/(3220000) = 0.0025 rad/s`

b) The radical acceleration:

`a_r = (v^2)/(R) = (8056^2)/(3220000) = 20.15 m/s^2`

c) Assume constant velocity, the tangential acceleration of the spaceship would be 0.