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The bus leaves the bus stop with an acceleration of 1.5 m.s^-2 which is mantained until it reaches the speed of 12 m.s^-1. At the same instant a girl who is 5 m away from the bus stop starts to run after the bus at a constant speed of 7 m.s^-1. Will the girl catch the bus? How should I link the two situations together mathematically? What would be the way?

The answer is that the girl will catch the bus and time would be 0.76 seconds. The girl is about 5 m away from the bus stop so would it be s=s-5? Key s=distance

Zayed
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  • As specified in the answer below, just use the general equations $v = v_0 + at$ and $x=x_0+u_0t+\frac{1}{2}at^2$ (assuming constant acceleration) with $u_0=0$ and $x_0$ specified accordingly. Also, assume that both the girl and the bus are moving in a parallel straight lines. – Winter Soldier Apr 16 '19 at 14:11

1 Answers1

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Time reference the moment the bus leaves and space reference the position of the bus when it leaves. Let $x_b$ the bus abscissa and $x_g$ the girl abscissa.

$x_g(t)=7t-5$ and $x_b(t)$ is more complicated to find. The bus has two type of motion; an accelerated one and an uniform one. Let’s find the starting of the uniforme phase. $v=\gamma t\Longrightarrow t=\frac{v}{\gamma}=\frac{12}{1,5}=8s.$ The bus has therefore these two equations: $$x_b(t)=0,75t^2\quad t\le 8$$ and $$x_b(t)=12(t-8)+0,75\times8^2=12t-48\quad t>8$$. Now solve $x_b(t)=x_g(t)$ with the time constraint i.e for $t\le8$ and $t>8$

For the first case i.e $t\le 8$ we solve $7t-5=0.75t^2\iff 0.75t^2-7t+5=0\iff t=8.55 \operatorname{ or } t=0.779$ since $t<8$ we choose $t=0.779$. The second case we solve $12t-48=7t-5$ which has one solution $t=\frac{43}{5}=8.6s$

Comment:

The first solution is the time that the girl catch the bus. If Amelia(the girl) decide to not jump in the bus and keep running she will exceed the bus but at $t=8.6s$ the bus will catch up Amelia.

DINEDINE
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  • Should the distance be s=(s-5) as she is 5m behind the bus stop or s=(s+5). And why? How do I choose what would be the correct one? I mean, on what basis should I choose to be the correct one? I mean I could solve for both s-5 and s+5 but.. you get my point! – Zayed Apr 18 '19 at 20:22
  • It depends what you choose as space reference. If the space reference is chosen at the girl position it will be +5 for the bus, if the space reference is at the bus position it gonna be -5 for the girl. It is specified in the answer above. – DINEDINE Apr 18 '19 at 23:07