When will the car and van collide if they do at all.

[~christina~]

Homework Statement

Speedy Sue during is driving at 30.0m/s
and enters a 1 lane tunnel. She then observes a slow moving van 155m ahead traveling at 5.0m/s. She applies the brakes but can only accelerate at -2.0m/s^2 b/c road is wet.

a) will there be a collision? How do you know?
b) If there is a collision state how far in the tunnel and at what time the collision occurs. If not then determine the distance of closest approach btwn the car and van.

sue:
vi= 30.0m/s
a= -2.0m/s^2
xi= 155m

Van:
vi= 5.0m/s




b]2. Homework Equations [/b]
not sure which kinematic eqzn to use...so many...

vxf=vxi +(ax)*t -----[velocity as a function of time]

xf= xi + 1/2(vxi + vxf)t----------[position as a function of velocity and time]

xf= xi + vxi*t +1/2 *ax*t^2-------------[position as a function of time

vxf^2= vxi^2 +2ax (xf-xi)---------[v as a function of position]

do I need xf= xi + vxt ? I guess that the answer would be no since no t is given but the van is going at a constant velocity of 5.00m/s
It shouldn't be used for Sue's car right? since the acceleration changes and velcocity too right since the car (sue) slows down and the velocity slows down too since my thing is that since a= v/t if a goes down then the v has to go down too.

The Attempt at a Solution

~well I know that
sue:
vi= 30.0m/s
a= -2.0m/s^2
xi= 155m

Van:
vi= 5.0m/s


I'm not sure what equation or what to do next. Is the acceleration constant? I assume so since this chap is about kinematics but it only says -2.0m/s^2 for acceleration so I guess it is..but do I need to find the xf for sue?

I think I would need to find xf but I don't have the final vf for Sue's car. Do I need to find that?


Thanks

 

[Astronuc]

In order to avoid the collision, Sue's car cannot over take the slower vehicle. What is the condition on the velocity (speed) of Sue's car to make that happen.

The other condition is the distance traveled - in the same time.

The slow vehicle will travel distance d and Sue's car must not exceed d+155m (the separation).

 

[m2003]

I would first clarify some information from the given problem. Is the tunnel a straight path or does it have any curves? Is there a specific length for the tunnel? These details can affect the outcome of the collision.

Assuming the tunnel is a straight path with a length of 155m, we can use the equations of motion to determine whether there will be a collision and at what point in time it will occur.

a) To determine if there will be a collision, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. For Sue's car, we have u = 30.0m/s, a = -2.0m/s^2, and s = 155m. Substituting these values, we get v^2 = 30.0^2 + 2(-2.0)(155) = 900 + (-620) = 280. This means that the final velocity of Sue's car will be √280 ≈ 16.73 m/s. Since the final velocity is positive, it means that the car will not stop before reaching the van. Therefore, there will be a collision.

b) To determine when and where the collision will occur, we can use the equations of motion again. We can use the equation v = u + at to find the time it takes for Sue's car to reach the van. We have v = 16.73 m/s, u = 30.0 m/s, and a = -2.0 m/s^2. Substituting these values, we get 16.73 = 30.0 + (-2.0)t. Solving for t, we get t ≈ 6.14 seconds.

To find the distance at which the collision occurs, we can use the equation s = ut + 1/2at^2. We have u = 30.0 m/s, t = 6.14 seconds, and a = -2.0 m/s^2. Substituting these values, we get s = 30.0(6.14) + 1/2(-2.0)(6.14)^2 = 184.2 m. This means that the collision will occur at approximately 184.2 meters into the tunnel