Catch the Bus or Frustration Distance?

[motoxtremechick]

A pedestrian is running at his maximum speed of 6.0 m/s to catch a bus stopped by a traffic light. When he is 25 meters from the bus, the light changes and the bus accelerates uniformly at 1.0 m/s2. Find either {A} how far he has to run to catch the bus or {B} his frustration distance (closest approach).

I have used d=Vf^2-Vi^2/2a and gotten 18 meters but I am not sure if I am correct..

 

[pooface]

I don't think this is correct. You can see that the pedestrian has to run more than 25 meters to catch up to the bus so 18 cannot be the answer.

I suggest you seperately write down your data for the pedestrian and the bus.

 

[motoxtremechick]

ok so
BUS
V=1.0m/s^2

PED
a=6.0m/s
d=25m

could i possibly use d=Vave(t) to find time and use 25 meters as distance or is that wrong?..

 

[pooface]

Bus has an acceleration of 1m/s^2, not velocity. It also has an initial velocity of 0 since it was at rest.

Pedestrian has a constant velocity of 6m/s not acceleration. The acceleration is 0 because V is constant. And if V is constant then Vo and Vf are the same.

you do not know the displacement, you only know that when the bus begins to acceleration the pedestrian is 25 metres behind. So it is a catch up game.

 

[motoxtremechick]

right right my bad on the A and V mess ups sorry.
and then would i use the d=Vf^2-Vi^2/2a to find a of the pedestrian?
putting 25m into d?

 

[pooface]

Well you can see if you use that formula you will get d=0/2a. Not what you want. Secondly 'a' of the pedestrian is 0 because he is running at a constant velocity.

you don't have displacement, so you can't use 25m.

I'll give you a headstart.
displacement of the pedestrian is
d=d+25
and displacement of the bus is d=d.
This is because the pedestrian is 25 meters behind the bus. If the bus displaced 'd' metres then the pedestrian must displace 'd +25m' at their meeting point if there is one.

Think about equating certain equations to find the time they will be at the SAME distance(d).

 

[motoxtremechick]

o alright.
thank you
im sorry I am just way lost and i guess not thinking straight
the only equations i have are the kinematics equations and i can't think of which one to apply that uses what i have/dont have

 

[pooface]

Analyze each equation, try to equate the d's. So effectively you will have to move the 25 over. Don't be afraid! I am here to help.
PED
d=d+25
Vo=6
Vf=6
a=0
t=?

BUS
d=d
Vo=0
Vf=?
a=1
t=?

More than enough information here

 

[motoxtremechick]

Thank you for bearing with me [=!
so just to get to the 25 meter mark from 0 the pedestrian would have to run 150 seconds...
wait

so i did d+25=Vf^2-Vi^2/2a?
so d+25=18m
d=-7 so he is 7m away from the bus? so 25 plus the extra 7 the bus travles so he is 32m away from the bus?

 

[pooface]

The pedestrian is running at 6 metres/second. To get to 25 metres, it will take him about 4.something seconds right? But during those 4.something seconds the bus is traveling forwards.

Try a different equation. You will need two equations for 'd'. One for the pedestrian and one for the bus.

Since 'd' is the same, you can equation the two equations and solve for something else.

example:
Bus: d=vot+0.5at^2
Ped: d+25=vot+0.5at^2

We don't have d, but we know they have to be the same.
vot+0.5at^2 = vot+0.5at^2 - 25

effectively we have eliminated d, and can solve for t.
Go back to the starting of 'example:' and see if you can simplify those equations with the data known.

 

[motoxtremechick]

ooo rigght..geez
hmm well i am very horrible at this
i don't know how to get t out of the equation effectively to solve for it..
would i just divide both sides by t^2 then...no see i am just digging a deeper hole with this problem

 

[pooface]

look back at the known data from post 8.

 

[motoxtremechick]

well
0m/s(t)+0.5(1.0m/s^2)(t^2)=6m/s(t)+0.5(0m/s^2)(t^2)-25
then by taking out the ones that would equal 0 out i reduced it to..
.5m/s^2(t^2)=6m/s(t)-25
hmm

 

[pooface]

cool looks like you have a quadratic!

What is t(the time the boy would catch up to the bus)?

 

[motoxtremechick]

i got t=-.064 and t=.3054
?

 

[pooface]

Equation is:

0.5t^2-6t+25 = 0

How did you get those values? Did you use the quadratic formula?

 

[motoxtremechick]

wait i think i switched my values
so i got
3.27 and -15.27

 

[pooface]

If I plug these values into the equation for t, I don't get zero.

Use the quadratic formula.

 

[motoxtremechick]

so
a=.5
b=-6
c=25

you can't have a - square root right
wouldnt those values give you a negative under the root sign?

 

[pooface]

Yes they would! So what happened here?

 

[motoxtremechick]

it couldn't have happened?

 

[pooface]

Exactly. This reveals that the pedestrian never caught up to the bus. Which leads to part B of the problem, what is the closest he gets? I guess that was the subliminal hint that he wouldn't make it.

 

[motoxtremechick]

haha oooo man ok wow
so now to figure out the closest..?? it wouldn't be the 25 meters would it?

 

[pooface]

I can tell you the bus driver makes a narrow escape.
hehehe.