What is the velocity, speed, position and distance traveled after 30 minutes?

[physics(L)10]

Homework Statement

A boat heading south initially with a speed of 10 m/s has a constant acceleration of a= (10,5,10) m/s^2

What is the velocity, speed, position and distance traveled after 30 minutes?

u=10m/s
a=(10,5,10)
t=30 min = 1800

Homework Equations

v = u + at

The Attempt at a Solution

a^2= 10^2 + 5^2 + 10^2
a=15

v= u + at
= 10 + 15x1800
=27010 m/s south

b) speed is just the scalar quantity so 27010 m/s (NOT SOUTH)

 

[kuruman]

I am not sure what kind of "boat" this is that has non-zero acceleration in all three directions. Assuming that it can fly, then you need to write the kinematic equations in the three directions independently, find each component after thirty minutes separately, then combine as needed using the Pythagorean theorem.

 

[physics(L)10]

Hahaha, sorry I ment a hot air balloon. Can you expand a bit more please, I still don't quite understand, sorry. I really need help because I'm in grade 12 and I would like to get into university, so I need a good mark in physics.

 

[kuruman]

Say "South" is x, "East" is y and "Up" is z. What expression will give you the x-component of velocity at any time t? What about the x-component of position at any time t?

If you can figure these out, then you should be able to apply the same reasoning to the other two directions. So start from there.

 

[xxChrisxx]

Also is this an acutal question set by someone? As a constant acceleation of 10m/s for 30 mins puts you beyond the escape velocity of earth.

You've got the right idea with the kinematic equations, but think about how you can describe the movement in 3 directions.

It's good practise to write out everything you know for each step, not just what you are given in the question. So for example each direction you have a:

Starting velocity
Acceleration
Staring Position.

You may have to define your own things, that aren't strictly necessary to solve the question but will help you to think through it step by step. So for example, it may be a good idea to define the start at coordinates (0,0,0).

 

[physics(L)10]

Say "South" is x, "East" is y and "Up" is z. What expression will give you the x-component of velocity at any time t? What about the x-component of position at any time t?

If you can figure these out, then you should be able to apply the same reasoning to the other two directions. So start from there.

So Vx=Ux + axt
Vy=Uy + ayt
Vz=Uz + azt

Where V=final velocity, U=initial velocity, a=acceleration and t=time

That seems about right, no? and then you use the other 4 kinematic equations in the same manner?

And chris, I don't know what that is lol. I think they just wanted you to apply the kinematic equations then anything else.

 

[xxChrisxx]

So Vx=Ux + axt
Vy=Uy + ayt
Vz=Uz + azt

Where V=final velocity, U=initial velocity, a=acceleration and t=time

That seems about right, no? and then you use the other 4 kinematic equations in the same manner?

And chris, I don't know what that is lol. I think they just wanted you to apply the kinematic equations then anything else.

Using those equations (and the displacemnt ones), do you know how to calcualte the final position?

I suspect that you have made the mental 'short cut' past what I've said, as it's fairly intutative, but I just find it easier to put every step down in full until I know I am sure what the steps are. It's just really hard to describe without giving the answer away.

 

[physics(L)10]

No, lol. I'm sorry, physics isn't my thing. I'm more of a chem guy.

 

[xxChrisxx]

Ok well can work through it from the very start.

Remember the path traveled can be done by vectors, so we can split the problem up into each axis.

For example if I walk 10 paces north and 10 paces west I end up in a certain place north west of my starting position. I can go north first or west first it doesn't matter.So the first thing to do is define the coordiantes where you start from.
You then work each direction out sepeately, as it doesn't matter which of the directions you travel first.

 

[physics(L)10]

Yeah I know that, that's very basic. It's the acceleration vector that's screwing me up because I don't know how to apply it to find velocity, speed e.t.c

 

[xxChrisxx]

Yeah I know that, that's very basic. It's the acceleration vector that's screwing me up because I don't know how to apply it to find velocity, speed e.t.c

Do you know how to calcualte the velocity, and distance in 1 dimension using the kinematic equations?

 

[physics(L)10]

If your given initial velocity, acceleration and time u can use v=u + at and then using that answer, you use s=(v+u/2)t. Correct?

 

[xxChrisxx]

If your given initial velocity, acceleration and time u can use v=u + at and then using that answer, you use s=(v+u/2)t. Correct?

Yep.

 

[physics(L)10]

The Attempt at a Solution

a^2= 10^2 + 5^2 + 10^2
a=15

v= u + at
= 10 + 15x1800
=27010 m/s south

b) speed is just the scalar quantity so 27010 m/s (NOT SOUTH)

so is this right?

 

[xxChrisxx]

so is this right?

Well yes, but you are skipping steps which is the reason you can't find the distance etc. Also the numbers are simply silly, that could be another reason it's not making sense to you.

You are trying to do it all as 1 vector, when you need to find the three components of the final traveled distance. That's why I talked about going through all the steps.

doing what you are doing gives you the right answers, but if you don't know why it's the right answer then you need to go through each step.

 

[physics(L)10]

Can you use this reasoning?

r=(x,y,z)
u=(x,y,z)
a=(10,510)

where r=initial position, u=speed, a=acceleration

Am I right here?

 

[xxChrisxx]

Can you use this reasoning?

r=(x,y,z)
u=(x,y,z)
a=(10,510)

where r=initial position, u=speed, a=acceleration

Am I right here?

Yeah, you've got the answer so we can talk about it now.

I assume you added the squares af the accelerations to find the resultant acceleration because of pythagoras? What you did there, what what I was going to do at the end. You found the resultant acceleration, I was going to find the resultsnats at the end.

If you did then, you got the right answer for the right reason.

 

[physics(L)10]

So 207010 is the right answer? and then to find distance you use s=(v+u/2)t which is s=(207010+10/2)1800 is your answer? But then I'm stuck on how to find position now.

Also, I have another question.

In my notes I suspect these equations having importance:

x=Xo + uxt + axt^2/2
y=Yo + uyt + ayt^2/2
z=Zo + uzt + azt^2/2

Does this relate to the equation s=ut + at^2/2? If so, what is the significance of Xo and x?

 

[xxChrisxx]

So 207010 is the right answer? and then to find distance you use s=(v+u/2)t which is s=(207010+10/2)1800 is your answer? But then I'm stuck on how to find position now.

Which is why you should have worked out the distance traveled in each component. you would have have 3 distances giving you the 3 coordiantes needed to define position.

Also, I have another question.

In my notes I suspect these equations having importance:

x=Xo + uxt + axt^2/2
y=Yo + uyt + ayt^2/2
z=Zo + uzt + azt^2/2

Does this relate to the equation s=ut + at^2/2?

Yes they are component form of the kinematic equation.

If so, what is the significance of Xo and x?

The X and X0 are a standard notation.

They subscript 0 means (position/velocity/acceleration) at time 0. The starting conditions in each axis x,y,z.

So where as you can use V and U for inital and final velocity you can use V0 and V.

 

[physics(L)10]

So that's what's screwing me up. How would you determine the distance traveled in each component?

And for some reason they substituted s for initial and final velocity in the equation s=ut + at^2/2?

 

[xxChrisxx]

So that's what's screwing me up. How would you determine the distance traveled in each component?

And for some reason they substituted s for initial and final velocity in the equation s=ut + at^2/2?

You do it the exact same way you did. You did 1 calcualtion for the resultant, you need to do 3 sets of calcualtions, 1 set for each component.

Then combine those results. So what you should have at the end is this.

Start position x,y,z
Start Velocity x,y,z
Starting acceleration x,y,z

Ending position x,y,z
Displacement x,y,z
Engine velocity x,y,z

From that you get:
Total displacement
Resultant velocity
SpeedI suggest you rework it following all the steps. Set your work out in a standard way. You also tehcnically need to define a coordinate system for it as it is it's meaningless without it. So you need to state which way is 'positive' for each axis.

 

[physics(L)10]

Okay, let's see if I can do it:

Let x=south, y=north, z=east

r=(0,0,0) because the initial position is not given, you assume
u=(10,0,0)
a=(10,5,10)

So now you use x=Xo + uxt + axt^2/2 which gives,
x=10+10(1800) + (15)(1800)^2/2

Am I on the right track? I feel like I'm doing it wrong

 

[xxChrisxx]

Okay, let's see if I can do it:

Let x=south, y=north, z=east

r=(0,0,0) because the initial position is not given, you assume
u=(10,0,0)
a=(10,5,10)

So now you use x=Xo + uxt + axt^2/2 which gives,
x=10+10(1800) + (15)(1800)^2/2

Am I on the right track? I feel like I'm doing it wrong

South and north are opposites. If you define +ve x as south, then -ve x is north.

 

[physics(L)10]

So this is basically a 2 dimension question because I can just put x=south, -x=north, y=west, -y=east?

 

[xxChrisxx]

So this is basically a 2 dimension question because I can just put x=south, -x=north, y=west, -y=east?

There's 3 dimensions. X Y and Z.

 

[physics(L)10]

Can you just do this question for me please. I'll understand it much better if I know exactly what to do.

 

[xxChrisxx]

We don't acutally do the questions for you here. I'll stick around to help you through it (got to leave work soon but i'll get on when I get home). It looks like you are getting confused beucase I've assumed a set level of knowledge. What level of maths have you achieved so far? (It'll help me better know how to explain things and where to start)

Do you know what a Cartesian coordinate system is?

 

[physics(L)10]

Lmao, I'm not that dumb man.

 

[xxChrisxx]

Thats a yes then. I'm only asking because you didn't seem to understand how the x y and z coordinate system works.

EDIT: I'm not trying to be offensive or anything, just trying to judge the level of mathematical knowledge.

 

[physics(L)10]

I understand that (z=up, y=toward you, x=across) I just don't understand how to apply it to the equations and which equations to use.

 

[xxChrisxx]

I understand that (z=up, y=toward you, x=across) I just don't understand how to apply it to the equations and which equations to use.

Well let's apply that to the situation. Forget the equations for the moment, and just think about what's going on.

You said it's a balloon,
so it can move north/south. We'll call that X.
With south in the +ve direction.

East/west we'll call that Y.
East in the +ve direction.

Up/down we'll call that Z.
Up is +ve.So the balloon can move in 3 different axis. In reality it can move in any direction. But that real movement can be defined as though you walked up/down, east/west, north/south.Do you know what resultants and components of something are?edit: I'm leaving work now. if you don't know what they are read the following:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
http://www.mathwarehouse.com/vectors/resultant-vector.php

 

[physics(L)10]

Isn't Up and down the same thing as north and south?

resultant is formed when you add/subtract two vectors together. components I probably know but I don't know they're called components so I don't know lol.

edit: component is a particular vector ex. (-1,10)?

 

[physics(L)10]

I've been working on it and I think I mightve gotten something:

r=(0,0,0)
u=(0,0,0)
a=(10,5,10)

and then you use the equation x=Xo +ut + at^2/2 for x,y,z and you get 3 velocity values. Now I'm stumped

 

[xxChrisxx]

Isn't Up and down the same thing as north and south?

Up and down is elevation.

resultant is formed when you add/subtract two vectors together. components I probably know but I don't know they're called components so I don't know lol.

and then you use the equation x=Xo +ut + at^2/2 for x,y,z and you get 3 velocity values. Now I'm stumped

Yep you should have 3 velocity values, and three displacement values.

Look at this graph. There are three pink lines, one going down x, along y and then up z. Those three that you've calculated correspond to those lines. Also you can see that the three displacements give you the new position (x,y,z).

What you now want to do is find the resultant of those three vectors. So that the path taken takes you straight from (0,0,0) to (x,y,z). #This is what you found originally.

 

[kuruman]

I've been working on it and I think I mightve gotten something:

r=(0,0,0)
u=(0,0,0)
a=(10,5,10)

and then you use the equation x=Xo +ut + at^2/2 for x,y,z and you get 3 velocity values. Now I'm stumped

What do the equations that you have written down represent? If you want to write the three vectors at time t = 0, you should write

r=(0,0,0)
u=(10,0,0)
a=(10,5,10)

Note that the x component of the initial velocity is 10 m/s as given by the problem. Now can you write the position, velocity and acceleration vectors not at just t = 0 but at any time t? Hint: The acceleration vector will be the same. What about the other two?