Why it's hard to go fast

We note that power is simply energy exerted over time. Since the formula for kinetic energy is $KE = \frac{1}{2} mv^2$ , we know that energy required in creating the force to counter drag is proportional to the square of the velocity of the object. Since the drag force is directly proportional to the velocity of the object, and the power is proportional to the force output, we have that the new power ratio is simply $1.2^2 =1.44$ .

We will use the formula $P=\frac{W}{\Delta t} = \frac{Fx}{\Delta t}=F \left(\frac{x}{\Delta t} \right) =Fv$

When the bicyclist travels at speed $v$ , the drag force is $-cv$ , so he needs to exert a force $F=cv$ to travel at constant speed. Thus, $P=Fv=cv^2$

Thus the power is directly proportional to the square of the speed. When the speed increases by $1.2$ times, the power increases by $1.2^2=1.44$ times. Therefore $\frac{P_1}{P_0} = \fbox{1.44}$ .

The general equation for Power is $\frac{dW}{dt}=\frac{d(F \times x)}{dt}=\frac{dF}{dt} \times x + \frac{dx}{dt} \times F$ , where F is the force and x is the displacement. The phrasing of the problem implies $\frac{dF}{dt}$ will be $0$ , because the problem asks for the power needed to maintain the new speed $1.2v$ , and taking the derivative of $F$ which will be dependent on a constant speed will be $0$ .

So, the Power at any speed will be $F \times \frac{dx}{dt} = F \times v =-cv^{2}$ . Plugging in $1.2v$ and $v$ and dividing, we get $1.44$

Power=Work/time.

Since Work=Force $\times$ distance, Power=Force $\times$ distance/time.

Since distance/time=speed, Power=Force $\times$ speed.

$F_0=-cv_0$ , so $P_0=F_0v_0=-cv_0^2$ .

$F_1=-cv_1$ , so $P_1=F_1v_1=-cv_1^2=-c(1.2v_0)^2$ .

$P_1/P_0=\boxed{1.44}$ .

$F_{drag} = -cv$

$P_{drag} = Fv = -cv^{2}$

when $P$ amount of power is applied , velocity increases and hence drag force also increases until $P_{applied} = - P_{drag}$

$P_{applied} = - (-cv^{2})$

$P_{applied} = cv^{2}$

Hence, $P_{applied}$ is directly proportional to $v^{2}$

$\frac{P_{1}}{P_{0}} = \frac{(1.2 v_{0})^{2}}{v_{0}^{2}}$

$\frac{P_{1}}{P_{0}} = 1.44$