Answer:
The shortest distance is [tex]S = 24.86 ft[/tex]
Explanation:
The free body diagram of this question is shown on the first uploaded image
From the question we are told that
The speed of the bicycle is [tex]v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s[/tex]
The distance between the axial is [tex]d = 42 \ in[/tex]
The mass center of the cyclist and the bicycle is [tex]m_c = 26 \ in[/tex] behind the front axle
The mass center of the cyclist and the bicycle is [tex]m_h = 40 \ in[/tex] above the ground
For the bicycle not to be thrown over the
Momentum about the back wheel must be zero so
[tex]\sum _B = 0[/tex]
=> [tex]mg (26) = ma(40)[/tex]
=> [tex]a = \frac{26}{40} g[/tex]
Here [tex]g = 32.2 ft/s^2[/tex]
So [tex]a = \frac{26}{40} (32.2)[/tex]
[tex]a = 20.93 ft/s^2[/tex]
Apply the equation of motion to this motion we have
[tex]v^2 = u^2 + 2as[/tex]
Where [tex]u = 32.26 ft /s[/tex]
and [tex]v = 0[/tex] since the bicycle is coming to a stop
[tex]v^2 = (32.26)^2 - 2(20.93) S[/tex]
=> [tex]S = 24.86 ft[/tex]
