Two particles, constant acceleration, when do they collide?

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Homework Help Overview

The problem involves two particles, a car and a van, moving with different velocities and accelerations. The scenario is set in a tunnel where the car is attempting to stop while the van continues at a constant speed. The main question is whether a collision occurs and, if so, the specifics of the collision in terms of distance and time.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations to determine the conditions for collision. There is confusion regarding the correct formulation of the equations and the variables involved, particularly concerning the final velocity and the initial conditions for both vehicles.

Discussion Status

Some participants have offered corrections to the equations being used and suggested methods for finding the time and distance of travel for both vehicles. However, there is still uncertainty among participants regarding the interpretation of the variables and the calculations needed to reach a conclusion.

Contextual Notes

There is mention of a graph being used to estimate the time of collision, but the original poster expresses being stumped by the problem setup. Additionally, there are indications of frustration and a request for further clarification on the equations being applied.

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Homework Statement



Speedy Sue, driving at 33.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 165 m ahead traveling with velocity 5.20 m/s. Sue applies her brakes but can accelerate only at -2.00 m/s2 because the road is wet.

Will there be a collision?

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 and enter zero for the time.

Homework Equations



Vx - Current velocity
V0 - Initial velocity
a - acceleration
x - Final Position
x0 - Initial displacement

vx^2 = v0+2a(x-x0)

The Attempt at a Solution


By doing a graph I estimated a crash at about 6-7 seconds but that is as far as I got...

I am completely stumpted on this "two particles" deal... I tried the above formula and somehow ended up with a 0 on the bottom because the van does not accelerate...

Any help / answer / example would be very much appreciated.

Thank you.
 
Last edited:
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vx^2 = v0+2a(x-x0)
This formula is wrong. It should be
vx^2 = vo^2 + 2a*(x - xo)
Using given values, find x. xo is zero.
Using vx = xo + at, find t.
When Sue and van meet, they must have traveled for the same time interval t.
Find the distance s traveled by the van is time t.
Compare x and s, and decide whether they just meet, collide or Sue falls behind.
 
Thank you very much.
 
rl.bhat said:
vx^2 = v0+2a(x-x0)
This formula is wrong. It should be
vx^2 = vo^2 + 2a*(x - xo)
Using given values, find x. xo is zero.
Using vx = xo + at, find t.
When Sue and van meet, they must have traveled for the same time interval t.
Find the distance s traveled by the van is time t.
Compare x and s, and decide whether they just meet, collide or Sue falls behind.

how do you know what vx^2 is in the bolded equation? thanks in advance

i've been staring at this problem for an hr...is vx^2, or velocity final, just 5.20 m/s? i don't know of any other way to derive v(final)...arg
 
Last edited:
anybody? :(
 
cc21392001 said:
anybody? :(
vx is the final velocity of Sue when she travels 165 m.
 

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