First Year Physics Help Needed

[TJ the Tornado]

This is my second day in physics class and we have merely learned the principles of displacement and velocity. I am having trouble using these principles in the problem which follows...


Runner A is initially 10.0 mi west of a flagpole and is running with a constant velocity of 8.0 mi/h due east. Runner B is initially 8.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when they meet?


I don't know where to begin so any help would be great thanks

 

[Tide]

I hope your instructor isn't giving you a trick question! (The runners will meet at the flagpole because each is knocked out when they collide with it!)

Seriously, what you need to do is to determine how far away from the flagpole each runner is at any given time. (Hint: the distance traveled by either one is his or her speed times the amount of time that has elapsed.) Once you have that you can determine how far apart they are from each other.

 

[Nenad]

to solve questions like these, you need to make use of some very simple kinematics equations, the two main one are:
$$v^2 = {v_0}^2 + 2ad$$

$$d = {v_0}t + \frac{1}{2}at^2$$

Hint, for this one you need to develp an equation for the runners distance from the flagpole as a function of time, and then equate and solve for a variable.

Regards,

Nenad

 

[Andrew Mason]

to solve questions like these, you need to make use of some very simple kinematics equations, the two main one are:
$$v^2 = {v_0}^2 + 2ad$$

$$d = {v_0}t + \frac{1}{2}at^2$$

Hint, for this one you need to develp an equation for the runners distance from the flagpole as a function of time, and then equate and solve for a variable.

?? I am not sure what all this relates to, but it does not appear to relate to this question.

The answer to the original question is simply a matter of determining the rate of change of separation, or relative speed, of the two runners. Just determine how long it takes until the separation distance = 0 and then work out where that point is.

AM

 

[mcah5]

x is displacement, v is velocity, t is time, x0 is initial displacement.

x = v*t + x0

Plug in v, x0 for both runner A and B, set them equal to each other and solve for t. That will give the time it takes for the runners to meet. Plug in t to either equation, and you get x.

 

[HallsofIvy]

Tide was, I think, joking. I would assume that the two runners are at least capable of stepping around the flag pole rather than running into it! (Of course, tide may have different experience with running into things than I do.)

Here's a very "formal" way of doing the problem. Set up a "coordinate system" ("number line" here) so that the flag pole is at 0 (since everything is given in terms of the flagpole), west is negative and east positive (that's arbitrary- just stick with one choice of positive and negative). Then A starts at -10 ("10 miles west of the flag pole). Since he runs at 8 mph (positive because he is running east) and distance run is "speed times time", his position after t hours is x(t)= -10+ 8t. B starts at +8 (8 miles east of the flagpole) and his speed is -5 mph (negative because he is running west), his position after t hours is x(t)= 8- 5t. They will meet when they are at the same positon at the same time: x(t)= -10+ 8t= 8- 5t. Solve that equation for t (the number of hours after starting that they meet) and plug into either of x(t)= -10+ 8t or 8- 5t to determine where they meet (miles east of the flagpole if x(t) is positive, miles west of the flagpole if x(t) is negative).

 

[Nenad]

?? I am not sure what all this relates to, but it does not appear to relate to this question.

The answer to the original question is simply a matter of determining the rate of change of separation, or relative speed, of the two runners. Just determine how long it takes until the separation distance = 0 and then work out where that point is.

AM

Andrew, I am not saying that he must use these equations in this question, I am telling him what to use in the future when encountering more difficult kinematics problems.