The time taken for you to catch up with the bus is 2.45 s
The maximum time you can wait before starting to run and still catch the bus is 0.84 s.
The given parameters include:
distance behind the bus door, d = 9 m
acceleration of the bus door, a = 1 m/s²
your initial velocity, u = 4.9 m/s
The time taken for you to catch up with the bus must occur at the exact position with the bus,
let the position = y
The time taken for you to reach this point:
[tex]t = \frac{total \ distance }{speed} = \frac{y + 9}{4.9} \ ---(1)[/tex]
The time taken for the bus to reach this same point;
[tex]y = vt + \frac{1}{2} at^2\\\\initial \ velocity \ of \ the \ bus = 0\\\\y = 0.5\times (1)t^2\\\\y = \frac{t^2}{2} \\\\t= \sqrt{2y} \ ---(2)[/tex]
solve equation (1) and (2) together;
[tex]\frac{y + 9}{4.9} = \sqrt{2y} \\\\(\frac{y + 9}{4.9} )^2= 2y\\\\\frac{y^2 + 18y + 81}{24.01} = 2y\\\\y^2 + 18y + 81 = 48.02y\\\\y^2 -30.02y + 81 = 0\\\\solve \ quadratic \ equation \ using \ formula \ method;\\\\a = 1 , b = -30.2 , c = 81\\\\y = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\y = \frac{-(-30.02) \ \ +/- \ \ \sqrt{(-30.02)^2 - 4(1\times 81)} }{2(1)}\\\\y = 2.998 \ m \ \ or \ 27.02 \ m[/tex]
The time taken for you to catch up is calculated with the smallest distance;
[tex]t = \frac{y + 9}{4.9} \\\\t = \frac{2.998 + 9}{4.9} = 2.45 \ s[/tex]
The maximum waiting time is calculated using relative velocity and acceleration:
[tex]a_r = - 1 \ m/s \\\\s = s_0 + \frac{1}{2} (-1) t^2 \\\\s = 9 -0.5t^2[/tex]
And equation for the relative velocity;
[tex]v_r = v_0 + a_rt\\\\v_r = 4.9 + (-1)t\\\\v_r = 4.9 -t[/tex]
The maximum waiting time is now calculated as follows;
[tex]v_f_r^2 = v_r + 2a_rs\\\\0 = (5-t) ^2 + 2 \times (-1) (9 - 0.5t^2)\\\\0 = (5-t )^2 -18 + t^2\\\\0 = 25 -10t + t^2 - 18 + t^2\\\\0 = 2t^2 -10t+ 7\\\\a = 2 , \ b = -10, c = 7\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 -4ac} }{2a} \\\\t = \frac{-(-10) \ \ +/- \ \ \sqrt{(-10)^2 -4(2\times 7)} }{2(2)} \\\\t = 4.15 \ s \ \ or \ \ 0.84 \ s[/tex]
Thus, the maximum waiting time is 0.84 s
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