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

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SUMMARY

In the scenario presented, Speedy Sue is driving at an initial speed of 30.0 m/s and applies brakes with a deceleration of -2.0 m/s² while approaching a van traveling at 5.0 m/s, positioned 155 meters ahead. To determine if a collision occurs, kinematic equations are utilized to calculate the stopping distance of Sue's car and the distance traveled by the van. The analysis concludes that if Sue's car does not exceed the distance traveled by the van plus the initial separation of 155 meters, a collision can be avoided.

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  • Kinematic equations for motion analysis
  • Understanding of acceleration and deceleration concepts
  • Basic algebra for solving equations
  • Knowledge of relative motion principles
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  • Calculate the stopping distance of a vehicle using the equation: xf = xi + vxi*t + 1/2 * ax * t²
  • Analyze the time taken for both vehicles to cover their respective distances
  • Explore the implications of constant acceleration on vehicle motion
  • Investigate collision avoidance strategies in vehicular dynamics
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Students studying physics, particularly in kinematics, automotive engineers, and anyone interested in vehicle dynamics and collision analysis.

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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:confused:
 
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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).
 

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