Distance traveled after a collision

[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.

 

[gneill]

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.

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

 

[kieslingrc]

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

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

 

[Screen]

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

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.

 

[kieslingrc]

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.

The distance is what I can't 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.

 

[gneill]

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

 

[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.

 

[gneill]

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.

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.

 

[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!

 

[Screen]

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!

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.