(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:

(b) The initial speed of the skater is

while the final speed is

and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by

(c) The initial speed of the skater is 

while the final speed is 

since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:

from which we find

where the negative sign means it is a deceleration.
 
        
        
        
Answer:
Coaches plan, conduct, and evaluate the team on a regular basis. They are the leaders of the team, so they help motivate the players. They are able to communicate with the athletes and determine the best learning styles for each player.
Explanation:
 
        
                    
             
        
        
        
The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N
 
        
             
        
        
        
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .
also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)