Question

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s2. You instantly start running toward the still-open door at 5.9 m/s. How long does it take for you to reach the open door and jump in? What is the maximum time you can wait before starting to run and still catch the bus?

Answer 1

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

$x_b(t) = d + \frac{1}{2}at^2$

where

d = 9.0 m is the initial distance between you and the bys

$a=1.0 m/s^2$ is the acceleration of the bus

Substituting,

$x_b(t)=9+0.5t^2$ (1)

The motion of the person is uniform, so its position at time t is

$x_p(t) = vt$

where

v = 5.9 m/s is the constant speed

Substituting,

$x_p(t)=5.9t$ (2)

The person reaches the bus when

$x_p(t)=x_b(t)\\5.9t=9+0.5t^2$

Re-arranging and solving for t,

$0.5t^2-5.9t+9=0$

$t=\frac{5.9\pm \sqrt{5.9^2-4(0.5)(9)}}{2(0.5)}=\frac{5.9 \pm 4.1}{1}=5.9\pm 4.1$ (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

$x_p(t')=5.9(t-t')=5.9t-5.9t'$

So the equation to solve becomes:

$x_p=x_b\\5.9t-5.9t'=9+0.5t^2\\0.5t^2-5.9t+(9+5.9t')=0$

This means that the discriminant of the solution of this second-order equation is

$\sqrt{(-5.9)^2-4(0.5)(9+5.9t')}$

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:

$(-5.9)^2-4(0.5)(9+5.9t')}=0$

And solving for t',

$(-5.9)^2-4(0.5)(9+5.9t')}=0\\34.8-18-11.8t'=0\\11.8t'=16.8\\t'=1.42 s$

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