Speed Average Velocity Change Question

[Kildars]

This is my first post here, I am going for high college physics, this is my second physics class in my life. (One in high school and this is Physics 100 -- Survey of Physics.)

We do all our homework online, so here's the first one..


A person walks first at a constant speed of 3.40 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 5.60 m/s. What is

(a) her average speed over the entire trip?

For average I simply did 3.4+5.6 = 9 Then I did 9/2 (Two different speeds) and I got 4.5.. but for some reason it's showing up wrong on my webassign. Is there something small I'm missing? Any help would be appreciated.


Second Question

A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 358 m in 168 s. What is her average velocity (both magnitude and direction) for the entire fall?
Magnitude
m/s

So for the free fall v = (a)(t) and because a = g in this instance I did v = (9.8)(15.0) = 147 m/s

So for the free fall velocity = 147

The second part is confusing, The parachute is open so a != 9.8 because of the wind drag on the chute, does a = 1/2gt^2? Cause if it does I think I can figure it out.

Thanks for your help in advance.

 

[Doc Al]

A person walks first at a constant speed of 3.40 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 5.60 m/s. What is

(a) her average speed over the entire trip?

For average I simply did 3.4+5.6 = 9 Then I did 9/2 (Two different speeds) and I got 4.5.. but for some reason it's showing up wrong on my webassign. Is there something small I'm missing? Any help would be appreciated.

Your thinking here is incorrect. Despite the temptation, you don't find average speed by adding two speeds and dividing by two (except under special conditions like constant acceleration). An exaggerated example will make the point: Say you travel one mile at a speed of 1 mile per year, then return at a speed of 1,000,000 miles per year. Is your average speed around 500,000 miles per year? No! You spend much more time going slower so your average speed is much slower!

Instead of trying to average the speeds directly, use the relationship: average speed = distance traveled divided by time.

Second Question

A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 358 m in 168 s. What is her average velocity (both magnitude and direction) for the entire fall?
Magnitude
m/s

So for the free fall v = (a)(t) and because a = g in this instance I did v = (9.8)(15.0) = 147 m/s

So for the free fall velocity = 147

The second part is confusing, The parachute is open so a != 9.8 because of the wind drag on the chute, does a = 1/2gt^2? Cause if it does I think I can figure it out.

Again, don't get hung up calculating velocities at various points. Use: average speed = distance traveled divided by time.

 

[Kildars]

Your thinking here is incorrect. Despite the temptation, you don't find average speed by adding two speeds and dividing by two (except under special conditions like constant acceleration). An exaggerated example will make the point: Say you travel one mile at a speed of 1 mile per year, then return at a speed of 1,000,000 miles per year. Is your average speed around 500,000 miles per year? No! You spend much more time going slower so your average speed is much slower!

Instead of trying to average the speeds directly, use the relationship: average speed = distance traveled divided by time.



Again, don't get hung up calculating velocities at various points. Use: average speed = distance traveled divided by time.

Ah, that's right because speed = velocity when the direction is the same. I forgot that point, alright let me work it out.

 

[Kildars]

Sweet, I got it.

a = d/t



s = 625m+358/(15+168)

625+358 = 983
15+168 = 183

987/183 = 5.37 == Right Answer.


For this one

A person walks first at a constant speed of 3.40 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 5.60 m/s. What is

(a) her average speed over the entire trip?

I don't know time traveled or distance traveled so how would I go about using savg = dt

 

[Kildars]

Anyone?

Also this question is bringing up problems also:

A rocket rises vertically, from rest, with an acceleration of 2.8 m/s^2 until it runs out of fuel at an altitude of 1300 m. After this point, its acceleration is that of gravity, downward. (Assume that the positive direction is upward.)

(a) What is the velocity of the rocket when it runs out of fuel?
m/s
(b) How long does it take to reach this point?
s
(c) What maximum altitude does the rocket reach?
m
(d) How much time (total) does it take to reach maximum altitude?
s
(e) With what velocity does it strike the earth?
m/s
(f) How long (total) is it in the air?
s

For A

I tried 1300/2.8 -- I got 464, I assumed that meant it took 464s to get to 1300m but v = at and (2.8)(464) = 1300 which obviously cancels them out, then I thought v = deltaX/deltaT deltaX being 1300 and deltaT being 464 and I got 20.31 which is also wrong..

I really want to do good in physics, I understand the formulas I just don't understand how to apply them to certain problems.

 

[radou]

Anyone?

Also this question is bringing up problems also:

A rocket rises vertically, from rest, with an acceleration of 2.8 m/s^2 until it runs out of fuel at an altitude of 1300 m. After this point, its acceleration is that of gravity, downward. (Assume that the positive direction is upward.)

(a) What is the velocity of the rocket when it runs out of fuel?
m/s
(b) How long does it take to reach this point?
s
(c) What maximum altitude does the rocket reach?
m
(d) How much time (total) does it take to reach maximum altitude?
s
(e) With what velocity does it strike the earth?
m/s
(f) How long (total) is it in the air?
s

For A

I tried 1300/2.8 -- I got 464, I assumed that meant it took 464s to get to 1300m but v = at and (2.8)(464) = 1300 which obviously cancels them out, then I thought v = deltaX/deltaT deltaX being 1300 and deltaT being 464 and I got 20.31 which is also wrong..

I really want to do good in physics, I understand the formulas I just don't understand how to apply them to certain problems.

Regarding B - use the expression for displacement of the rocket to obtain the time it takes to reach 1300 meters. Further on, plug that time into the equation of the velocity of the rocket. You now have the answer to A, too. For the other parts, use the equation of displacement for a free fall - meaning the only acceleration is gravity.

 

[Kildars]

Regarding B - use the expression for displacement of the rocket to obtain the time it takes to reach 1300 meters. Further on, plug that time into the equation of the velocity of the rocket. You now have the answer to A, too. For the other parts, use the equation of displacement for a free fall - meaning the only acceleration is gravity.

I think what I am most confused on is how to find time? Is the time to 1300m really 464s? Or did I do that wrong?

 

[radou]

I think what I am most confused on is how to find time? Is the time to 1300m really 464s? Or did I do that wrong?

You got it wrong. The equation of displacement equals x(t) = 1/2*a*t^2 . Plug in the acceleration and height into the equation and solve for t.

 

[Kildars]

You got it wrong. The equation of displacement equals x(t) = 1/2*a*t^2 . Plug in the acceleration and height into the equation and solve for t.

That doesn't make sense to me, where do I plug in "height" I see where to plug in acceleration, but t^2 is not height.. unless you meant x(d) as in change in distance..

 

[radou]

That doesn't make sense to me, where do I plug in "height" I see where to plug in acceleration, but t^2 is not height.. unless you meant x(d) as in change in distance..

x(t) means: the height x in the moment t (i.e. the change of distance with respect to time). So, if we write 1600 = 1/2*2.8*t^2, we get the moment t from the equation, which is the moment the rocket reaches the height of 1600 meters.

 

[Kildars]

x(t) means: the height x in the moment t (i.e. the change of distance with respect to time). So, if we write 1600 = 1/2*2.8*t^2, we get the moment t from the equation, which is the moment the rocket reaches the height of 1600 meters.

Well I figured out a and b

A is 85.32
B is 30.47

I got A by doing:

$$v^2 = 2(a) \Delta X$$
2a = (2.8)(2) which equals 5.6
(5.6)(1300) = 7280
Square Root(7280) = 85.32

I got B by doing:

85.32 = 0 + (2.8)t
divide by 2.8
You get 30.47
so t = 30.47

so C

(c) What maximum altitude does the rocket reach?
m

When it runs out of fuel at 1300m the acceleration becomes -9.8m/s and at that point the velocity is 85.32. So I tried to do:
$$v^2 = v_0^2 + 2 a \Delta x$$

0 = 85.32 + 2*-9.8(x)
-85.32 = -19.6(x)
which is 4.35
and 4.32^2 is 18.94 which is incorrect.

So what did I do wrong on C?

 

[radou]

Just use $$v = v_{0} + at$$, i.e. 0 = 85.32 - 9.81 * t , and solve for t to get the time it takes to reach the maximum altitide. Then, plug that time t into $$x = x_{0} + v_{0}t + \frac{1}{2}at^2$$, i.e. x = 1300 + 85.32*t + 1/2*(-9.81)*t^2 in order to get the maximum height.

 

[Kildars]

Just use $$v = v_{0} + at$$, i.e. 0 = 85.32 - 9.81 * t , and solve for t to get the time it takes to reach the maximum altitide. Then, plug that time t into $$x = x_{0} + v_{0}t + \frac{1}{2}at^2$$, i.e. x = 1300 + 85.32*t + 1/2*(-9.81)*t^2 in order to get the maximum height.

Okay, so

-85.32/9.81 = 8.697

So

$$x = 1300 + 85.32*8.697 + 1/2*(-9.81)*8.697^2$$

$$8.697^2 = 75.642$$

$$x = 1300 + 85.32*8.697 + 1/2*(-9.81)*75.642$$

Then 85.32 * 8.697 = 742.028
Then 1/2(-9.81) = -4.905
Then -4.905 * 75.642 = -371.024
So x = 1300 + 742.021 + 371.024
So x = 1670.997 which is the correct answer, thank you.

 

[Kildars]

Anyone?

Also this question is bringing up problems also:

A rocket rises vertically, from rest, with an acceleration of 2.8 m/s^2 until it runs out of fuel at an altitude of 1300 m. After this point, its acceleration is that of gravity, downward. (Assume that the positive direction is upward.)

(a) What is the velocity of the rocket when it runs out of fuel?
m/s
(b) How long does it take to reach this point?
s
(c) What maximum altitude does the rocket reach?
m
(d) How much time (total) does it take to reach maximum altitude?
s
(e) With what velocity does it strike the earth?
m/s
(f) How long (total) is it in the air?
s

For E.

How would I start E, I know the answers of A, B, C, D.

A: 85.32m/s
B: 30.47s
C: 1670.997m
D: 38.47s
E: ?

(e) With what velocity does it strike the earth?
m/s

I tried $$v^2 = v_0^2 + 2 a \Delta x$$
$$v^2=85.32+2(-9.8) \Delta x$$
v^2 = 65.72(1670.997)
v^2 = 109817.922
v = 331.38 m/s

Which is incorrect, since I don't know the time it takes from 1670-0 I can't use $$v = v_0 + a t$$.

Any help or advice would be appreciated.

 

[Hootenanny]

At the rocket's highest point what is it's velocity (it's not 85.32 m/s)?

 

[Kildars]

Using v = vi + at

I got v = 85.32 + (-9.8)*(8.7)
v = .06

Is that correct?

 

[Hootenanny]

Using v = vi + at

I got v = 85.32 + (-9.8)*(8.7)
v = .06

Is that correct?

Not quite. What does the rocket do once it reaches it's highest point?

 

[Kildars]

Not quite. What does the rocket do once it reaches it's highest point?

It stops, the velocity is 0. I was sitting in class a realized it, haha. I feel stupid ;).

Anyways, I figured it out it was -180.91 m/s. :)

Thanks for your help I'll probably need help on some other stuff so if I can't figure it out I'll post ;)

 

[Hootenanny]

It stops, the velocity is 0. I was sitting in class a realized it, haha. I feel stupid ;).

Anyways, I figured it out it was -180.91 m/s. :)

Thanks for your help I'll probably need help on some other stuff so if I can't figure it out I'll post ;)

One gets used to that feeling, unfortunately my eureka moments usually happen just as I'm falling asleep :grumpy:

 

[Kildars]

For this one



I don't know time traveled or distance traveled so how would I go about using savg = dt

For some reason this problem is really getting past me, I've tried a couple different things and still can't figure it out. I can't do savg = d divided by time because I know neither.

 

[Doc Al]

You don't need to know the distance. Try this: Call the distance from A to B "D". Compute the total time and distance in terms of the speeds and D and see what happens.

 

[Kildars]

You don't need to know the distance. Try this: Call the distance from A to B "D". Compute the total time and distance in terms of the speeds and D and see what happens.

I know that is supposed to help me, but I don't understand what that means, How can I use any of the formulas that I know at the moment, with two unknowns? I don't know acceleration, distance, or time.

 

[Doc Al]

I know that is supposed to help me, but I don't understand what that means, How can I use any of the formulas that I know at the moment, with two unknowns? I don't know acceleration, distance, or time.

You have the distance: it's D! Now use the speed to find the time from A to B, and from B to A. Then find the average speed for the entire trip. Try it!

 

[Kildars]

You have the distance: it's D! Now use the speed to find the time from A to B, and from B to A. Then find the average speed for the entire trip. Try it!

t = 3.4d
t = 5.6d

so savg = d/t

I still don't understand how this helps me because I have two unknowns.

 

[Doc Al]

For one thing, Speed = Distance/Time. So redo your time calculations.

Then find:
(1) Total time
(2) Total distance

Then, finally, calculate the average speed. Ave Speed = Total Distance/Total Time.

 

[Kildars]

For one thing, Speed = Distance/Time. So redo your time calculations.

Then find:
(1) Total time
(2) Total distance

Then, finally, calculate the average speed. Ave Speed = Total Distance/Total Time.

5.6 = d/t1
3.4 = d/t2

t1 + t2 = ttotal


Again, how do I find time total with two unknowns? I'm sorry if I seem really stupid right now, I just know that I can't find something when I have two unknowns and even though I know distance = "D" "D" is still a variable..

 

[Doc Al]

First rule: Do what Doc Al tells you! (Just kidding.)

Find t1. Write it down here:

Find t2. Write it down here:

Add them up. That's the total time. Just do it.

D is the one way distance. What's the round trip distance traveled?

 

[Kildars]

First rule: Do what Doc Al tells you! (Just kidding.)

Find t1. Write it down here:

Find t2. Write it down here:

Add them up. That's the total time. Just do it.

D is the one way distance. What's the round trip distance traveled?

Find t1. Write it down here: 5.6t1 = d

Find t2. Write it down here: 3.4t2 = d

I don't understand how to add them up with two variables, sorry :(.

There's got to be another way.

 

[Doc Al]

Find t1. Write it down here: 5.6t1 = d

t1 = d/5.6

Find t2. Write it down here: 3.4t2 = d

t2 = d/3.4

I don't understand how to add them up with two variables, sorry :(.

Just do it!
t1 + t2 = d/5.6 + d/3.4 = d(1/5.6 + 1/3.4)

That's the total time. Now what's the total distance (round trip)?

 

[Kildars]

t1 = d/5.6


t2 = d/3.4


Just do it!
t1 + t2 = d/5.6 + d/3.4 = d(1/5.6 + 1/3.4)

That's the total time. Now what's the total distance (round trip)?

d = (5.6/d1 + 3.4/d2)

?

 

[Doc Al]

Nope, it's much simpler than that. If "d" is the one way distance, what's the round trip distance?

 

[Kildars]

Nope, it's much simpler than that. If "d" is the one way distance, what's the round trip distance?

2d ;) .

1234567890

 

[Doc Al]

Yes (finally!)

Now use the total time and distance to find the average speed.

 

[Kildars]

Well I did d(1/5.6 + 1/3.4)

I got d(.472).. but I don't know a number for distance so I can't multiply that out..

:(

 

[Doc Al]

You don't need a number for the distance (or the time for that matter). Take the expression for total distance that you found and divide it by the expression for total time.