# Distance traveled after a collision

1. Dec 14, 2011

### kieslingrc

1. A 1000 kg aircraft going 25 m/s collides with a 1500 kg aircraft that is parked and they stick together after the collision and are going 10 m/s after the collision. If they skid for 13.2seconds before stopping, how far did they skid? Hint: Are the aircraft moving at a constant velocity after the collision or do they experience an acceleration?

2. distance(d) = 1/2acceleration(a)*time(t)^2;

3. I am lost on this one. I am working with linear momentum equations right now but only know how to solve for the speed, not the distance.

2. Dec 14, 2011

### Staff: Mentor

If they skid for 13.2 seconds, what's their velocity after 13.2 seconds? Did the velocity change over the 13.2 seconds?

3. Dec 14, 2011

### kieslingrc

Yes, they came to a rest so the velocity changed from 10 m/s to 0

4. Dec 14, 2011

### Screen

That means you know the initial velocity and you know the final velocity, and the time it took to stop - what can you find from that? Try listing all the variables you know.

5. Dec 14, 2011

### kieslingrc

The distance is what I cant find. It's not a constant velocity. Here is what I know:
total mass (m) = 1500kg
the change in velocity is -10
the change in time is 13.2

I have the equation to calculate the speed of the aircraft before or after the collision, but they were both given variables. I have no equation in my text that I could find to use that combination of variables in solving for distance, hence the reason I am stuck.

6. Dec 14, 2011

### Staff: Mentor

You have the change in velocity and the time. What's the acceleration?

7. Dec 14, 2011

### kieslingrc

acceleration = the change in velocity dived by the change in time. So, -10 divided by 13.2 gives me -.75757575... which I would round up to -.76

distance = 1/2 at^2 so (.5 * -.76)(13.2)^2; -.38 * (13.2)^2; -.38 * 174.24 = -66.21

This just doesn't seem right.

8. Dec 14, 2011

### Staff: Mentor

The acceleration looks good. But you're only using part of the kinematic equation for the distance. There's an initial velocity to consider, too.

9. Dec 14, 2011

### kieslingrc

So I just said to heck with it and put that in. WRONG! here is what it say's the formula and answer is:

In this case, the two aircraft go from 10 to 0 m/s in the amount of time given, which means the acceleration a = (-10 m/s) divided by the time "t". This means we have a negative acceleration. However, the initial velocity vo = 10 m/a and

Vf = V0 + at or 0 = 10 + at or a = -10/t and d = v0t + .5 at2 = 10t +(.5)(-10/t)(t2) = 10t - 5t

d = 5 t (m)

Sure is nice when your taking an online class and none of the text or videos they provide even remotely resembles this!

Last edited: Dec 14, 2011
10. Dec 14, 2011

### Screen

If you care the problem had nothing to do to with momentum (well, you would use momentum to find the final velocity, but that was already given to you - 10 m/s), instead it was just one dimensional motion.

If it starts at 10 m/s, and it takes 13.2 seconds to stop, the acceleration would be

$a =$$\frac{v}{t}$, $\frac{10}{13.2}=$ -.758 $m/s^2$ (negative because it's coming to a stop).

In words, every second that passes the velocity decreases by -.758, therefore, it would reach a stop (0 m/s) at 13.2 seconds.

Now that you know the acceleration, just plug your numbers into the formula:

$x_f = x_o + v_ot + .5at^{2}$

$x_o$ is equal to zero because that's where it starts, $v_o$ is 10, because that is it's initial velocity, $t$ is 13.2, that's the time it stops and, therefore, the time of which it achieves its farthest displacement (it can't move anymore), and $a$ is -.758, which you just solved.

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