a) 7 buses
b) 6.8 m
Explanation:
a)
The motion of the motorcycle is a projectile motion, so it consists of 2 independent motions:
- A uniform motion along the horizontal direction
- A uniformly accelerated motion along the vertical direction
The initial components of the velocity of the motorcycle are:
[tex]v_x = u cos(32^{\circ})=(40.0)(cos 32^{\circ})=33.9 m/s\\v_y = u sin(32^{\circ})=(40.0)(sin 32^{\circ})=21.2 m/s[/tex]
The equation for the vertical motion of the motorcycle is
[tex]y=h+u_y t - \frac{1}{2}gt^2[/tex]
where
y is the altitude at time t
h is the initial height
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
The bus top is at the same height of the initial ramp, so we have
[tex]y=h[/tex]
And therefore, we can solve the equation for t, to find the time of flight:
[tex]0=u_y t - \frac{1}{2}gt^2\\t(u_y-\frac{1}{2}gt)=0\\t=\frac{2u_y}{g}=\frac{2(21.2)}{9.8}=4.33 s[/tex]
Now we find what is the horizontal distance covered by the motorcycle in its jump, which is given by:
[tex]d=v_x t = (33.9)(4.33)=146.8 m[/tex]
And since each bus has a length of L = 20.0 m, the number of buses that the motorcycle can clear with its jump is:
[tex]n=\frac{d}{L}=\frac{146.8}{20}=7.34[/tex]
So, 7 buses.
b)
In the previous problem, we saw that the total range of the motion of the motorcycle is
[tex]d=146.8 m[/tex]
And we said that this corresponds to 7 buses.
Each bus has a length of
L = 20 m
So, the total length of 7 buses is
[tex]L' = 7L=7(20)=140 m[/tex]
Therefore, the range of the motorcycle is greater than the length of the buses by:
[tex]\Delta x = d-L'=146.8-140 = 6.8 m[/tex]
which means he will miss the last bus by 6.8 meters.
