Question

On a windless day a cyclist goes an incline uphill at a constant speed
of 12 km/hr and downhill on same incline at 16 km/hr. Air resistance is
proportional to square of his speed and his maximal effort which he
applies on pedal which causes friction on tyre to move the cycle is
independent of speed. His maximum speed on his maximal effort on a
flat road is close to :-
(A) 20 km/hr
(B) 18 km/hr
(C) 22 km/hr
(D) 14 km/hr

Answer 1

Answer:

22km ha brother goodmornimg

Answer 2

Answer:

The maximum speed on his maximal effort on a flat road is close to option (d) 14km/hr.

Explanation:

Step 1:

The speed of the cyclist going uphill $v_u = 12 km/hr$

The speed of the cyclist going downhill $v_d = 16 km/hr$

Let the mass of the cyclist is m, angle of the inclined surface is θ.

Let the maximum force that the cyclist applies on the pedal is F and the friction due to the surface is f.

air drag ∝ $(speed)^2$

⇒air drag = $k(speed)^2$

where k is the proportionality constant.

The horizontal forces for all three cases uphill, downhill, and flat roads are shown in the figure below.

Step 2:

The cyclist moves with a constant velocity in all three cases therefore the acceleration is zero.

The equation of motion for the uphill motion (see figure):

$m g \sin \theta+f-F+k v_{u}{ }^{2}=0$

The equation of motion for the downhill motion (see figure):

$m g \sin \theta+F-f-k v_{d}^{2}=0$

Subtracting equation (2) from (1) we get,

$2 f-2 F+k v_{u}{ }^{2}+k v_{d}^{2}=0$

On rearranging we get,

$F-f=\frac{k\left(v_{u}{ }^{2}+v_{d}^{2}\right)}{2}$

The equation of motion for motion on a flat road (see figure):

$F-f-k v^{2}=0$

On rearranging,

$F-f=k v^{2}$

Step 3:

Using equation (3) and (5) we get,

$F-f=k v^{2}=\frac{k\left(v_{u}{ }^{2}+v_{d}{ }^{2}\right)}{2}$

$\Rightarrow v^{2}=\frac{v_{u}{ }^{2}+v_{d}{ }^{2}}{2}$

$\Rightarrow v=\sqrt{\frac{v_{u}{ }^{2}+v_{d}^{2}}{2}}$

On substituting, $v_{u}=12 \mathrm{~km} / \mathrm{hr}$  and $v_{d}=16{~km} / {hr}$ we get,

$v &=\sqrt{\frac{(12 \mathrm{~km} / \mathrm{hr})^{2}+(16 \mathrm{~km} / \mathrm{hr})^{2}}{2}}$

$&=\sqrt{\frac{144+256}{2}} \mathrm{~km} / \mathrm{hr}$

$&=\sqrt{200} \mathrm{~km} / \mathrm{hr}$

$&=14.14 \mathrm{~km} / \mathrm{hr}$

Therefore the maximum speed on the cyclist's minimal effort on a flat road is close to 14 km/hr.