Kinematics Problem: Cockroach Collision Calculation and Solution

[chonny]

Please help me...I've been struggling for a long time over this following problem...

Some cockroaches can run as fast as 1.5 m/s. Suppose that two cockroaches are separated by a distance of 60 cm and that they begin to run toward each other at the same moment. Both insects have constant acceleration until they meet. The first cockroach has an acceleration of 0.20 m/s^2 in one direction and the second one has an acceleration of 0.12 m/s^2 in the opposite direction. How much time passes before the two insects bump into each other?


Please give me the answer and the steps leading to the answer as soon as possible...I am most grateful

If not...I will kill myself

 

[liljediboi]

first convert the cm to m so you use the same units. then use your logic. if two things are accelerating towards each other, then what is their acceleration?

after, just plug it into the formula

and on a side note, this belongs in homework help

 

[chonny]

Well...I'm stuck at this certain point...

Roach #1 has the A=.20 m/s^2
Well Time = Change in Velocity / Acceleration
1.5m/s dividedby .2m/s^2 = 7.5 s

Roach 1 would have a displacement of 5.625m from 1/2 * 1.5m/s * 7.5s

Roach #2 has the A=.12 m/s^2
1.5m/s dividedby .12m/s^2 = 12.55s

Roach 2 would then have a displacement of 9.375m from 1/2 * 1.5m/s * 12.55s

...now I'm stuck...how would I figure out where they would bump in that .6m distance and how much time passes when they bump

 

[HallsofIvy]

Originally posted by chonny
Well...I'm stuck at this certain point...

Roach #1 has the A=.20 m/s^2
Well Time = Change in Velocity / Acceleration
1.5m/s dividedby .2m/s^2 = 7.5 s

Roach 1 would have a displacement of 5.625m from 1/2 * 1.5m/s * 7.5s

Roach #2 has the A=.12 m/s^2
1.5m/s dividedby .12m/s^2 = 12.55s

Roach 2 would then have a displacement of 9.375m from 1/2 * 1.5m/s * 12.55s

...now I'm stuck...how would I figure out where they would bump in that .6m distance and how much time passes when they bump

You appear to be assuming that the cockroaches accelerate to 1.5 m/s and that they meet when they reach that speed. That is not said anywhere in the problem. It simply says that "some cockroaches can run 1.5 m/s". Other than suggesting that the speed should not be more than that, it really isn't relevant to the problem.

If the first roach has acceleration 0.2 m/s^2, then its speed at any time, t, is 0.2 t m/s and the distance traveled (displacement) is 0.1 t^2 m.

If the second roach has acceleration .12 m/s^2, then its speed at any time, t, is 0.06 t m/s and the distance it traveled is 0.03 t^2 m.

At the time they meet they will, together, have covered the entire
60 cm= 0.6 m distance between them:

0.1 t^2+ 0.03 t^2= 0.6

Solve that for t to find when they meet and use that to determine the distance each has run to find where they meet.

Roach