Two particles, constant acceleration, when do they collide?

In summary: So vx = 33 m/s and vo = 5.20 m/s. The acceleration a is -2.00 m/s^2. Plug in these values into the equation and solve for x.
  • #1
denxnis
6
0

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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  • #2
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
Thank you very much.
 
  • #4
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:
  • #5
anybody? :(
 
  • #6
cc21392001 said:
anybody? :(
vx is the final velocity of Sue when she travels 165 m.
 

1. What is the formula for calculating when two particles with constant acceleration will collide?

The formula for calculating when two particles with constant acceleration will collide is t = (v0 - u0) / a, where t is the time of collision, v0 and u0 are the initial velocities of the particles, and a is the constant acceleration.

2. How do the initial velocities of the particles affect the time of collision?

The initial velocities of the particles directly affect the time of collision. The larger the difference between the initial velocities, the shorter the time of collision will be.

3. Can the particles collide at more than one point?

No, two particles with constant acceleration can only collide at one point. This is because the paths of the particles are determined by their initial velocities and accelerations, and these values are constant.

4. What happens if the particles have different accelerations?

If the particles have different accelerations, they will not collide. This is because their paths will not intersect at any point, and they will continue moving in different directions.

5. Is the formula for calculating the time of collision applicable to all types of constant acceleration?

Yes, the formula for calculating the time of collision is applicable to all types of constant acceleration, including both positive and negative accelerations. The only requirement is that the acceleration remains constant throughout the particles' motion.

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