Answer :
a) You can catch the bus after 1.8 s or 10 s
b) The maximum time that you can wait is 1.42 s
Explanation:
a)
The motion of the bus is an accelerated motion, so its position at time t can be written as
[tex]x_b(t) = d + \frac{1}{2}at^2[/tex]
where
d = 9.0 m is the initial distance between you and the bys
[tex]a=1.0 m/s^2[/tex] is the acceleration of the bus
Substituting,
[tex]x_b(t)=9+0.5t^2[/tex] (1)
The motion of the person is uniform, so its position at time t is
[tex]x_p(t) = vt[/tex]
where
v = 5.9 m/s is the constant speed
Substituting,
[tex]x_p(t)=5.9t[/tex] (2)
The person reaches the bus when
[tex]x_p(t)=x_b(t)\\5.9t=9+0.5t^2[/tex]
Re-arranging and solving for t,
[tex]0.5t^2-5.9t+9=0[/tex]
[tex]t=\frac{5.9\pm \sqrt{5.9^2-4(0.5)(9)}}{2(0.5)}=\frac{5.9 \pm 4.1}{1}=5.9\pm 4.1[/tex] (3)
Which gives two solutions:
t = 1.8 s
t = 10 s
b)
If the person waits a time t' before starting to run, then the position of the person is given by
[tex]x_p(t')=5.9(t-t')=5.9t-5.9t'[/tex]
So the equation to solve becomes:
[tex]x_p=x_b\\5.9t-5.9t'=9+0.5t^2\\0.5t^2-5.9t+(9+5.9t')=0[/tex]
This means that the discriminant of the solution of this second-order equation is
[tex]\sqrt{(-5.9)^2-4(0.5)(9+5.9t')}[/tex]
In order for the person to still catch the bus, there must be at least 1 real solution to the equation, so the discrimant must be at least zero:
[tex](-5.9)^2-4(0.5)(9+5.9t')}=0[/tex]
And solving for t',
[tex](-5.9)^2-4(0.5)(9+5.9t')}=0\\34.8-18-11.8t'=0\\11.8t'=16.8\\t'=1.42 s[/tex]
Learn more about accelerated motion:
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