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~christina~
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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