Will Speedy Sue collide with the van in the tunnel?

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Homework Statement


Speedy sue, driving at 30.0m/s enters a one lane tunnel. She then observes a slow-moving van 155m ahead traveling at 5.00m/s. Sue applied her breaks and accelerates at -2.00 m/s^2. Will there be a collision? If yes, at what distance and what time?


Homework Equations


initial v = 30 m/s
final v = 0
a = -2/s^2
displacement = [(final velocity - initial velocity) / 2*acceleration)]

The Attempt at a Solution


By using the above equation, I found out that a crash does occur when Sue stops. However, I do not understand how to find out at exactly what displacement this crash occurs.
 
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The position of the truck at a time t is given by xtruck(t)=155m+t*(5m/sec). I've picked x=0 to be the tunnel entrance and t=0 to be the time when Sue enters the tunnel. Can you write an equation for Sue's position at time t? Then set them equal and solve for t.
 
Start by expressing Sue's displacement as a function of time and the truck's displacement as a function of time. At some time they end up in the same spot at the same time. (Measure the displacement of both from Sue's initial position.)
 
I'm sorry, I don't understand. How can I express sue's, or the truck's displacement as a function of time?
 
saber1357 said:
How can I express sue's, or the truck's displacement as a function of time?
Dick gave you the truck's position as a function of time. Now you find Sue's position as a function of time.
 
I don't understand how Dick came up with that function :)
I want to say that the function for Sue's car is car(t) = t*30m/s, but that doesn't work out because I don't know how to incorporate the acceleration. Could you please explain how a function of time is created?
 
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You must have some kinematical equations to use. The one I'm thinking of looks like:

x(t)=x0+v0*t+(1/2)*a*t^2.

Does that look familiar?
 
aha! so the equation for the car will be car(t) = 30t + .5(-2 m/s^2)(t^2)?