# Two particles, constant acceleration, when do they collide?

1. Sep 12, 2009

### denxnis

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

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

3. 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: Sep 12, 2009
2. Sep 12, 2009

### rl.bhat

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.

3. Sep 13, 2009

### denxnis

Thank you very much.

4. Sep 26, 2009

### cc21392001

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: Sep 26, 2009
5. Sep 27, 2009

### cc21392001

anybody? :(

6. Sep 27, 2009

### rl.bhat

vx is the final velocity of Sue when she travels 165 m.