# Girl running after a bus, kinematic equations.

1. Sep 27, 2015

### LogarithmLuke

1. The problem statement, all variables and given/known data
Marie is running to catch a bus. The bus has stopped at a bus stop. The bus has the acceleration 0.75m/s^2. Marie has the speed 8m/s. Before the bus starts driving, the space between Marie and the bus is 17.5m.
A How long will it take for Marie to catch up to the bus?
B What distance did she have to run to catch up to the bus?
C If Marie keeps running, at some point the bus will catch up to her. How long will this take?

2. Relevant equations
All the 1-D kinematic equations. I'm not able to write the equations like that so i think it is easier if i just link this.

3. The attempt at a solution
A

So my idea was to find out how long it took for Marie to run the distance 17.5m, and then add the time it would take for her to run the distance the bus has driven in this time. I don't think this is right, because for the time it takes for her to run the extra distance the bus drove when she ran the intital 17.5m, the bus will have driven even further. I just cant figure out any other ways to solve it. I first found the time it took for Marie to run 17.5m, which is time is equal to distance divided by speed, 17.5m/8m/s=2,1875. I used 2.2 as the time for the rest of the problem. The next thing i did was to find V(end speed) after 2.2s for the bus, using the following equation: V=v0+at
V=0+0.75*2.2=1,65m/s After this i tried to find the distance the bus drove over this time. Here i used the following kinematic equation: S=1/2(v0+v)*t i put in my numbers: S=1/2(0+1.65m/s)*2.2=1.815m I now add this to the intitial 17.5m which Marie ran and get 19.315. I now divide this by 8 and get 2.4. And i get that it took Marie 2.4s to catch up to the bus. Like i said earlier i do not think this is correct, but that was the way i attempted to solve it.

B

This question is what really got me to doubt my way of solving a. I already found the whole distance, so what is the point of this B task then? It made me think that finding the whole distance first was not the way to go.

C

I am interpreting this like I have to find when the bus has the speed 8m/s, because that is when it will be driving as fast as Marie is running. 0.75x=8 x=10,6s. It took the bus 10.6s to catch up to Marie. My way of solving this one seems way too simple, and 10 secounds seems like way too long aswell.

2. Sep 27, 2015

### Staff: Mentor

In (A) you know that Marie and the bus each travel for the same duration before their paths coincide, so call this time "t" and determine their respective distances of travel in terms of t.

See how you go with that.

3. Oct 8, 2015

### LogarithmLuke

I still haven't been quite able to figure out this one. I can't really find a good equation to use the common t in.

4. Oct 8, 2015

### Matejxx1

try it using this