Help with Physics Question?

Question

1. A person walks first at a constant speed of 5.23 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 3.10 m/s. (a) What is the average speed over the entire trip? (b)What is the average velocity over the entire trip?

2. On your wedding day your lover gives you a gold ring of mass 3.86 g. 53 years later its mass is 3.17 g. On the average how many atoms were abrated from the ring during each second of your marriage? The atomic mass of gold is 197 u.

3. A tortoise can run with a speed of 10.8 cm/s, and a hare can run 20.7 times as fast. In a race, they both start at the same time, but the hare stops to rest for 115 seconds. The tortoise wins by a shell (19.4 cm). (a) How long does the race take? (b)What is the length of the race?

4. A driver in a car traveling at a speed of 62.2 mi/h sees a deer 115 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).

5. A jet plane lands with a velocity of +115 m/s and can accelerate at a maximum rate of -4.84 m/s2 as it comes to rest. (a) From the instant it touches the runway, what is the minimum time needed before it can come to rest? (b) Can this plane land on a small tropical island airport where the runway is 1296.13 m long?

Answer 1

uh...do your own homework.

you're not going to learn anything if we just tell you.

Just use the kinematics equations

Answer 2

these are a lot of questions to answer, let me try to get you started on some of them...

1) Question one deals with the difference between speed and velocity.  Speed is the total distance covered divided by the total time covered; sometimes it is useful to use a specific distance (even though one isn't given).

Let's say the distance between A and B is 100 m.  Then, the first leg of the trip takes a time of 100m/5.23 m/s or 19.1s.  The return trip takes 100m/3.1 m/s = 32.3 s

We have that the total trip is 200 m and takes a total time of 51.4 s.  The average speed for the total trip is 200m/51.4s =3.89 m/s (notice this is NOT the same you would get from averaging 5.23 and 3.1)

For the average velocity, remember that velocity is displacement (a vector) /time.  Since the person begins and ends at the same place, the displacement is zero and the vector average velocity is also zero.

5) The plane's speed upon touching the landing strip is 115m/s and the plane slows at the rate of 4.84 m/s/s.  Thus, it will take 115m/s / 4.84 m/s/s = 23.8 s for the plane to come to rest.

The equation I think you are most likely to have equating distance traveled with speed and acceleration is:

dist = v0xt + 1/2 at^2

where v0 is the initial speed (115m/s here), t is the time of travel (23.8s) and a is the accel (-4.84 m/s/s)

the distance needed for the plane to come to rest is:

dist = 115x23.8 - 1/2 (4.84)(23.8)^2

dist = 2737-1371=1366m

so the plane will overshoot the runway.