A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is shown in the figure above. The value for each force is shown in the figure. What is the acceleration of the student-bicycle system?
m
А
0
2
B
0.2"
C) 5
D
25"

Solution:

The gravitational force that acts on the bicycle system is

Now the force, that is the gravitational force is related to mass of the system and the acceleration due to gravity of the system, 'm' and 'g' respectively.

Therefore, we can write

500  = m x 10    (since , g = 10 m/s-s)

∴ m = 50 kg

Now the net vertical force acting on the student bicycle system is 0. And the vertical acceleration of system is also 0. The total horizontal force acts to the right of the system. So by Newton's 2nd law of motion, we can write

 

Therefore

Hence (C) is correct option.  

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.