Physics student in need of some help

[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.

 

[KingNothing]

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

 

[m2003]

Hello,

As a physicist, I understand that linear motion problems can be challenging, especially when dealing with multiple variables. I can offer some guidance to help you solve this problem.

First, let's define the variables in this problem. The cockroach's speed is given as 1.50m/s. The man's speed is 0.80m/s. The initial distance between them is 0.90m. We are also given the distance the cockroach needs to travel (1.20m) and we need to find the minimum acceleration needed for the man to catch up with the cockroach.

Now, let's think about the motion of both the cockroach and the man. The cockroach is moving away from the man at a constant speed of 1.50m/s. The man is moving towards the cockroach at a constant speed of 0.80m/s. This means that the cockroach's speed is always greater than the man's speed, so the man will never be able to catch up with the cockroach if they continue at these speeds.

To catch up with the cockroach, the man needs to accelerate. This means that his speed will increase over time. We can use the formula: d = ut + (1/2)at^2 to solve this problem, where d is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

We know that the man starts 0.90m behind the cockroach, and we need to catch up with it when it has traveled 1.20m. This means the distance traveled by the man is 1.20m - 0.90m = 0.30m. The initial velocity of the man is 0.80m/s. We can plug these values into the formula and solve for the time it takes the man to catch up with the cockroach.

0.30m = (0.80m/s)t + (1/2)a(t^2)

We also know that the cockroach is moving at a constant speed of 1.50m/s, so the distance it travels in the same time t is 1.50t. We can set this equal to the distance traveled by the man (0.30m) and solve for t.

1.50t = 0.30m

t = 0.20s

Now that we have the time,