Large cockroaches can attain a speed of 1.70 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 1.70 m/s as you move toward it at a speed of 0.730 m/s .
If you start a distance 0.990 behind it, what minimum constant acceleration would you need to catch up with it when it has traveled a distance 2.00 , just short of safety under a counter?
Thanks in advance guys :D
2007-01-28 18:09:34 · 4 answers · asked by erica s 1 in Science & Mathematics ➔ Physics
Let's see...we'll go with this, A is acceleration, V is velocity, P is position. T is time, in seconds. We'll set 0 as the roach's start position. You can use known information and derivitives to determine equations for A V and P.
Cockroach
A=0
V=1.7
P=1.7T
You
A=X (unknown variable)
V=XT+.73
P=(X/2)T^2 +.73T - .99
The finishing point is 2.00, and you want to figure out what X is, in order to arrive at the same time as the roach, which will arrive there at T=2/1.7, or 1.18. So we fill in 1.18 for T, and set P at 2, for your P equation.
2=(x/2)1.18^2 + .73*1.18 - .99
2=(x/2)1.39 + .86 - .99
2=(x/2)1.39 - .13
2.13=(x/2)1.39
1.53=x/2
x=3.06
You need to accelerate at a minimum of 3.06 m/s^2
2007-01-28 18:20:14 · answer #1 · answered by Master Maverick 6 · 1⤊ 0⤋
Large Cockroaches
2016-11-17 00:41:21 · answer #2 · answered by charleston 4 · 0⤊ 0⤋
Are u really asking a question or do need help with ur home work? {//_+"} Emoz rock!
2007-01-28 18:19:49 · answer #3 · answered by stephanie 2 · 0⤊ 0⤋
use a gun and you wont have to do the math lol
2007-01-28 18:24:16 · answer #4 · answered by BajaRick 5 · 1⤊ 0⤋