Kinematics

[saber1357]

1. The problem statement, all variables and given/known data
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?


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

3. 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.

[Dick]

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.

[Doc Al]

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.)

[saber1357]

I'm sorry, I don't understand. How can I express sue's, or the truck's displacement as a function of time?

[Doc Al]

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.

[saber1357]

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?

[Dick]

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?

[Doc Al]

You need to be familiar with the basic kinematic equations for constant speed and accelerated motion. Here's a summary that might prove helpful: Basic Equations of 1-D Kinematics (http://www.physicsforums.com/showpost.php?p=905663&postcount=2)

[saber1357]

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

[Doc Al]

Exactly!

[Dick]

Absolutely right.

[saber1357]

i love you all