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 5.7 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

Answer:

$t=1.893\ s$

$t=1.27\ s$ 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, $d=9\ m$

• acceleration of the bus, $a'=1\ m.s^{-2}$

• speed of the passenger, $v=5.7\ m.s^{-1}$

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:

$t=\frac{9+x}{5.7}$ .....................(1)

Time taken by the bus to reach this point:

$x=u'.t+\frac{1}{2} a'.t^2$

Since bus starts from rest, u'=0

$t=\sqrt{2x}$ ...................(2)

From (1) & (2)

$9+x=5.7\times \sqrt{2x}$

$81+x^2+18x=64.98x$

$x^2-46.98x+81=0$

$x=45.187\ m\ or\ 1.792\ m$

Now we take the smallest distance: x=1.792 m

Time taken for both to reach this point: (put x in eq. (1))

$t=\frac{1.792+9}{5.7}$

$t=1.893\ s$

Now for the maximum waiting time we find have:

• relative acceleration, $a_r=-1\ m.s^{-2}$ since the bus is moving away from the intended

• the final relative velocity for the passenger to catch it, $v_{r}=0\ m.s^{-1}$

Using equation of motion:

$s=9+\frac{1}{2} \times (-1)t^2$

also the initial relative velocity:

$u_r=v+a_r.t$

$u_r=5.7-t$

and

$v_r^2=u_r^2+2a_r.s$

$0^2=(5.7-1)^2-2\times1\times (9+0.5t^2)$

$t=1.27\ s$