Respuesta :
Answer:
[tex]t=1.893\ s[/tex]
[tex]t=1.27\ s[/tex] The time which the passenger can delay and still catch the bus must be less than this.
Explanation:
Given:
- distance between the door of the bus and the passenger, [tex]d=9\ m[/tex]
- acceleration of the bus, [tex]a'=1\ m.s^{-2}[/tex]
- speed of the passenger, [tex]v=5.7\ m.s^{-1}[/tex]
Now in order for the passenger to get to the door of the bus the time taken by the bus and the passenger to reach a common point must be equal.
Let them meet after x distance from the starting point of the bus.
Time taken by the passenger to reach this point:
[tex]t=\frac{9+x}{5.7}[/tex] .....................(1)
Time taken by the bus to reach this point:
[tex]x=u'.t+\frac{1}{2} a'.t^2[/tex]
Since bus starts from rest, u'=0
[tex]t=\sqrt{2x}[/tex] ...................(2)
From (1) & (2)
[tex]9+x=5.7\times \sqrt{2x}[/tex]
[tex]81+x^2+18x=64.98x[/tex]
[tex]x^2-46.98x+81=0[/tex]
[tex]x=45.187\ m\ or\ 1.792\ m[/tex]
Now we take the smallest distance: x=1.792 m
Time taken for both to reach this point: (put x in eq. (1))
[tex]t=\frac{1.792+9}{5.7}[/tex]
[tex]t=1.893\ s[/tex]
Now for the maximum waiting time we find have:
- relative acceleration, [tex]a_r=-1\ m.s^{-2}[/tex] since the bus is moving away from the intended
- the final relative velocity for the passenger to catch it, [tex]v_{r}=0\ m.s^{-1}[/tex]
Using equation of motion:
[tex]s=9+\frac{1}{2} \times (-1)t^2[/tex]
also the initial relative velocity:
[tex]u_r=v+a_r.t[/tex]
[tex]u_r=5.7-t[/tex]
and
[tex]v_r^2=u_r^2+2a_r.s[/tex]
[tex]0^2=(5.7-1)^2-2\times1\times (9+0.5t^2)[/tex]
[tex]t=1.27\ s[/tex]
