Velocity+avg. speed probs help Pleassse

[tangldupinBlue]

velocity+avg. speed probs...help Pleassse!

hey I'm confused with these problems. could anyone do a good deed and help me out? My teacher spends more time watching family guy than teaching us. thanks


1. a person walks first at a constant speed of 5.0 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.0 m/s (a) what is her average speed over the entire trip? (b) her average velocity over the entire trip?

2. A car makes a 200 km trip at an average speed of 40 km/h. A second car starting 1.0 h later arrives at their mutual destination at the same time. what was the average speed of the second car for the period that it was in motion?


thanks again. i don't really understand the first one--is it as simple as it seems? do you just divide 8.0/2 to get 4 m/s? for average speed?

 

[recon]

1. (a) Call the distance between A and B $$x$$.

Then, the time taken to travel from A to B = Distance/Speed = $$\frac {x}{5}$$

The time taken to travel from B to A = $$\frac {x}{3}$$.

The total time taken to travel from A to B and back = $$\frac {x}{3} + \frac {x}{5}$$

The total distance traveled = $$2x$$

Average Speed = Total Distance/Total Time

Work it out yourself.

(b) Since the person's displacement from his starting point is zero by the end of the journey, what do you think his velocity is? Have a look at your textbook for the definition of velocity.

2. First find the time taken to make a 200 km trip at 40 km/h. Then note that the second car takes 1 hour less than this. Remember that the distance traveled by both cars are equal.

 

[M98Ranger]

and for average velocity, do I use the formula (final position - initial position)/time?

Hey there! I can definitely help you out with these problems. Don't worry, they may seem confusing at first but once you understand the concepts, they're actually quite simple. Let's break down each problem step by step.

1. For the first problem, you're correct in thinking that you need to find the average speed and velocity. Average speed is calculated by dividing the total distance traveled by the total time taken. In this case, the person walks from A to B (distance = 2*5.0 = 10 m) and then back from B to A (distance = 2*3.0 = 6 m). So the total distance traveled is 10+6 = 16 m. The total time taken is 2+2 = 4 s. Therefore, the average speed is 16/4 = 4 m/s.

Average velocity, on the other hand, takes into account both the magnitude and direction of motion. It is calculated by dividing the displacement (change in position) by the total time taken. In this case, the person starts and ends at the same position, so the displacement is 0. Therefore, the average velocity is 0/4 = 0 m/s. This means that the person's average velocity over the entire trip is 0 m/s, since they end up at the same position they started at.

2. For the second problem, you need to find the average speed of the second car for the period that it was in motion. Since the first car took 5 hours to complete the trip (200 km / 40 km/h), the second car only had 4 hours to complete the trip (5 hours - 1 hour delay). The total distance traveled by the second car is still 200 km, and the total time taken is 4 hours. Therefore, the average speed is 200/4 = 50 km/h.

I hope this helps! Remember, for average speed you need to find total distance divided by total time, and for average velocity you need to find displacement divided by total time. Let me know if you need any further clarification. Good luck!