# Physics student in need of some help

• nyguy
In summary, the problem is asking for the minimum constant acceleration needed for a person to catch up with a cockroach that is moving away from them at a constant speed of 1.50m/s. The person is initially moving towards the cockroach at 0.80m/s and starts 0.90m behind the cockroach. The solution involves using the equation for distance and acceleration, with the distance being the sum of the initial distance between the person and the cockroach and the distance the person needs to travel to catch up. The initial velocity of the person is 0.8 m/s and the acceleration is the unknown variable.

#### nyguy

I'm fairly new here and I was in search of some help to clarify some problems. I know that it may sound very easy to most on this forum but I seem to be having a tough time with straight-line (linear) motion problems. When being asked specific questions in a problem I notice that one has to go beyond what's originally being asked of you. For example, I might be asked to find the acceleration, but I might need to find the time to help solve the puzzle. Also, problems dealing with the acceleration of two bodies can really get to me as well. Here is an example of a problem that I took straight out of my textbook. Please help:

Large cockroaches can run as fast as 1.50m/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 1.50m/s as you move toward it at 0.80m/s. If you start 0.90m behind it, what minimum constant acceleration would you need to catch up with it when it has traveled 1.20m, just short of safety under a counter?

I was having problems determing if 0.80m/s was the man's initial velocity or the final velocity when he turned on the light. Also, 0.90m threw me off a bit. Please help and thanks in advance.

0.8 m/s is the man's initial velocity, it seems to be a straightforward application of the equation relating distance and acceleration:
$$$d = v_0 + {\textstyle{1 \over 2}}at^2$$$

-t will be the time it takes the cockroach to reach its shelter
-distance will be the distance the person needs to travel (.9m+1.2m)
-initial velocity is 0.8 m/s