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Homework Statement
35. Speedy Sue, driving at 30.0 m/s,
enters a one-lane tunnel. She then
observes a slow‐moving van 155 m
ahead traveling at 5.00 m/s. Sue
applies her brakes but can accelerate
only at ‐2.00m/s2 because the road is
wet. Will there be a collision? State
how you decide. If yes, determine
how far into the tunnel and at what
time the collision occurs. If no,
determine the distance of closest
approach between Sue’s car and the
van.
Homework Equations
s = v0(t1 - t0) + 1/2a(t1 - t0)^2
The Attempt at a Solution
I have reached an answer to this problem, but I am not sure that it is correct.
I'll copy my notes as I wrote them:
We need to find displacement in the amount of time decelerating at -2.00m/s^2.
v0 = 30m/s
a = -2.00m/s^2
So, I need time to solve for the displacement, so I used t=v/a to get 15seconds. This was (30m/s)/(-2.00m/s^2) which gave me -15s. I am not sure if I went astray here because I interpreted the -15s as the time it would take for the vehicle to stop.
I plugged this all into the displacement equation to get s=225m
So, if Suzie is going 30m/s and decelerating at -2.00m/s^2, she will not stop until 225m after 15 seconds.THE VAN
The van would need to travel 70+ meters in 15 seconds in order to avoid collision. I came to this conclusion since the van is already 155m ahead of Suzie and she ends up at 225m.
The velocity of the van is 5m/s, so in 15 seconds, the van will travel 75 meters. Adding this distance to the distance between Suzie's vehicle and the van gives us 230 meters which is clear of the 225 meters that Suzie's vehicle will travel.
Distance of closest approach
I am not entirely sure about this one, but I would imagine that the distance of closest approach would be when Suzie's vehicle comes to a halt at 225m. This would make the closest approach 5m.Any help is very much appreciated.
Thanks :)