Answer:
F_d,1 / F_d,o = 4 , Option A
Explanation:
Given:
- The velocity of the bike V_o is doubled to 2*V_o
Find:
what happens to the magnitude of the drag force?
Solution:
- The expression of the Drag force as a function of objects area, coefficient of drag, velocity and density of medium is given by:
F_d = 0.5*C_d*A*p*V^2 / 2
- In our case the force of drag with speed V_o is:
F_d,o = 0.5*C_d*A*p*V_o^2 / 2
- force of drag with speed 2V_o is:
F_d,1 = 2*C_d*A*p*V_o^2 / 2
- Take the ratio of F_d,1 to F_d,o:
F_d,1 / F_d,o = (2*C_d*A*p*V_o^2 / 2) / (0.5*C_d*A*p*V_o^2 / 2)
We get: F_d,1 / F_d,o = 4
Hence, by doubling the speed the magnitude of drag force is increased by a factor of 4.
