Two particles, constant acceleration, when do they collide?

AI Thread Summary
Speedy Sue, traveling at 33.0 m/s, is approaching a slow-moving van at 5.20 m/s, positioned 165 m ahead. Sue's braking capability is limited to an acceleration of -2.00 m/s² due to wet road conditions. The discussion revolves around determining if a collision occurs and calculating the time and distance of closest approach if it does not. Participants clarify the correct application of the kinematic equation, emphasizing that both Sue and the van must be analyzed over the same time interval to assess their positions. The conversation highlights the confusion surrounding the final velocity and the need for accurate calculations to resolve the problem.
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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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