more acceleration

• Sep 10th 2007, 07:13 PM
cinder
more acceleration
Okay, this one I've looked at and I don't think I'll figure it out before someone posts. :)

Large cockroaches can attain a speed of 2.20 m/s in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant speed 2.20 m/s as you move toward it at a speed of 0.800 m/s.

If you start a distance 0.980 m behind it, what minimum constant acceleration would you need to catch up with it when it has traveled a distance 2.10 m, just short of safety under a counter?

I think since I'm not given a time, I'll have to use the equation that doesn't use it (obvious... er, right?).
• Sep 10th 2007, 07:36 PM
topsquark
Quote:

Originally Posted by cinder
Okay, this one I've looked at and I don't think I'll figure it out before someone posts. :)

Large cockroaches can attain a speed of 2.20 m/s in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant speed 2.20 m/s as you move toward it at a speed of 0.800 m/s.

If you start a distance 0.980 m behind it, what minimum constant acceleration would you need to catch up with it when it has traveled a distance 2.10 m, just short of safety under a counter?

I think since I'm not given a time, I'll have to use the equation that doesn't use it (obvious... er, right?).

Call the point where you start the origin and the direction of travel of the cockroach to be +x.

You've got to compare two charts of information. The first chart is the cockroach:
It starts at $x_0 = 0.980~m$ and travels a constant speed of $v = 2.20~m/s$, so it's position vs. time equation is
$x = x_0 + vt = 0.980 + 2.20t$

You know you want to intercept the cockroach when it has traveled 2.10 m, so this is at $x = 0.980~m + 2.10~m = 3.08~m$

So
$3.08 = 0.980 + 2.20t \implies t = \frac{3.08~m - 0.980~m}{2.20~m/s} = 0.954545~s$

So you have to intercept the cockroach at a position of 3.08 m in 0.954545 s.

You start at the origin, $x_0 = 0~m$ with a speed of $v_0 = 0.800 m/s$. We are assuming that you might be accelerating, so your position vs. time equation is:
$x = x_0 + v_0t + \frac{1}{2}at^2$
and you need to be at $x = 3.08~m$ at $t = 0.954545~s$.

Thus
$3.08 = 0 + 0.800(0.954545) + \frac{1}{2}a(0.954545)^2 \implies a = 6.59302~m/s^2$

-Dan
• Sep 10th 2007, 07:43 PM
cinder
Okay, following your work I end up with $a = 5.14 m/s^2$.
• Sep 10th 2007, 08:04 PM
topsquark
Quote:

Originally Posted by cinder
Okay, following your work I end up with $a = 5.14 m/s^2$.

I did make a boo-boo when I plugged into my calculator. (0.08 is not quite the same as 0.8!) However, I'm now getting 5.08 m/s^2. Tolerably close, but there may be an error in your work somewhere as well. (Unless you didn't do the whole problem at once and you did it in steps, rounding as you went... That would account for the discrepency.)

-Dan