Answer:
a.After 
second Mr Comer's speed 
b.Distance travelled by Mr.Comer in seconds 
Explanation:
a. Lets recall our first equation of motion 
Now we know that 
, 
and
Plugging the values we have.
Then Mr.Comer's speed after sec 
b.
Lets find the distance and recall our third equation of motion.
So 
distance covered.
Dividing both sides with 2a we have.
Plugging the values.
So Mr.Comer will travel a distance of .
Given: Mass of earth Me = 5.98 x 10²⁴ Kg
Radius of earth r = 6.37 x 10⁶ m
G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: Smallest possible period T = ?
Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r
but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²
GMe = 4π²r³/T²
T² = 4π²r³/GMe
T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 25,563,909.77 s²
T = 5,056.08 seconds or around 1.4 Hour
Magnitude of acceleration = (change in speed) / (time for the change).
Change in speed = (27 - 0) = 27 m/s
Time for the change = 10 s
Magnitude of acceleration = (27 m/s) / (10 s) = 2.7 m/s² .
