Bike Approaching Stopped Truck at Traffic Lights

[AltruistKnight]

Homework Statement

A truck is stopped at a traffic light. The moment the light turns green, it begins to accelerate constantly at 3.1m/s^2. A cyclist approaching the truck from behind with a constant velocity of 10m/s was 7.5m behind the truck when the lights turned green.

1)How much time elapses between the light turning green and the bike passing the truck?
2)How long does it take for the truck to re-pass the bike?
3)How far has the truck traveled from the lights when it re-passes the bike?

Homework Equations

This is a Kinematics question, so all the Kinematic Equations are applicable, which are as follows:
[tex]d=vit+1/2at^2[/tex]
[tex]vf=vi+at[/tex]
[tex]vf^2=vi^2+2ad[/tex]

The Attempt at a Solution

When the one passes the other, they would both have to be at the same position. I made their initial positions as "x" for the truck, and "x-7.5" for the bike. As the bike will gain on and pass the truck until they have the same speed, I know that the speed they will have when the bike first passes the truck will be less than or equal to the bike's initial speed. Unfortunately, I'm having difficulty seeing how the Kinematic Equations can get me started here. I get the feeling it'll involve some sort of quadratic, but I'm unsure of how to properly make the two aforementioned starting points of the truck and bike into a quadratic form, and from there how to derive additional values for my three Kinematic Equations.

 

[Delphi51]

The truck has accelerated motion so your d=Vit+1/2at² applies
What formula applies for the distance function of the bike in motion without acceleration?
Once you have your two d = formulas, taking the 7.5 m into account, you will solve by writing
truck distance = bike distance (maybe position would be a better term than distance)
then replace the two distances with your two formulas and solve the quadratic for t.

 

[AltruistKnight]

How would the 7.5 end up modifying the bike's equation? (I'm using d=Vit for it, I suspect it'd end up being Vit+7.5, since the bike's displacement would require it clear the 7.5m gap and as it reaches the final displacement where the truck is)

 

[Delphi51]

I wrote d = -7.5 + 10t for the bike so that at time zero it is 7.5 m behind from the truck's initial position.
It is way easier to write two formulas for the bikes position and the truck's position than to leap to the difference between them.

 

[AltruistKnight]

Ah, I see. -7.5 since the total distance the bike travels will be 7.5 more than that the truck traveled when they attain the same overall position. Thanks a lot, this problem was giving me a lot of grief.

 

[Delphi51]

The idea is to capture the motion of the bike in the formula. When you see d = -7.5 + 10t, think
t=0, d=-.7.5
t=1, d= 2.5 (past where the back of the truck was a second ago!)
t =2, d = 12.5
Now if you capture the truck's motion, too, you can work out where they are relative to each other. Formulas that tell everything about the motion as functions of time are tremendously powerful.
Another approach would be to graph them both on a d vs t graph. Of course these two equations are just the equations of the line for the bike and the parabola for the truck.