Find an expression of acceleration

In summary, the particle has a velocity which is given as: v = 3.2i+4.3j+5.4k m/s. It's acted by force which is given as: f = 7.1i+8.2j+9.3k.
  • #1
ibysaiyan
442
0

Homework Statement


A particle has mass m = 0.1 kg , has a velocity which is given as: v = 3.2i+4.3j+5.4k m/s
It's acted by force f = 7.1i+8.2j+9.3k

The questions which arise are:
i) Find the speed of the particle.
ii) Write down a unit vector in the direction of the velocity.
iii)Find the angle between the projection of the velocity in the x,y plane and the x - axis.
iv) Find an expression of acceleration.
v) Find the velocity at a time of 2s.

Homework Equations


Now the equations that come to my mind are:
F = ma and f = change of momentum / time


The Attempt at a Solution



To get the speed: wouldn't that be the same as finding out an r.m.s value as such particle speed becomes: ~7.6m/s ?
For part (ii)is this how we find out a unit vector ? (ii)[itex]\vec u / ||\vec u||[/itex]
Now I'm a little lost about part (iii). You see I know how to find an angle between two vectors but here, I suppose since it's speaking of x,y axis within the same vector so I will use the following equation: A .B = |A| x |B| cos theta Can someone please clarify this for me?

Part iv doesn't seem that tricky... expression for acceleration has to be: a = f/m ...
v) Finding out velocity could be find out by using f = momentum /time...


Some background information: I have just started my course in Astrophysics, and this question comes out of a tutorial sheet where they don't expect us to know much and just see how much knowledge we have retained.(Proper lessons haven't even started, however I want to finish this early)


P.S: I'd like to apologize in advance of not using proper notation or over the lack of latex usage.


-ibysaiyan
 
Physics news on Phys.org
  • #2


Anyone ?
 
  • #3


Why in the world would you want to trick your teacher into thinking you know things you don't? You say that "this question comes out of a tutorial sheet where they don't expect us to know much and just see how much knowledge we have retained." If you cannot do this problem yourself, then your teacher needs to know that so he/she can teach it!
 
  • #4


ibysaiyan said:
Anyone ?
Patience is a virtue. Your question has only been up for a few hours.
ibysaiyan said:
Now I'm a little lost about part (iii). You see I know how to find an angle between two vectors but here, I suppose since it's speaking of x,y axis within the same vector so I will use the following equation: A .B = |A| x |B| cos theta Can someone please clarify this for me?
You have a vector in [itex]\mathbb{R}^3[/itex] and you want to find the angle between its projection onto the xy-plane and the x-axis. In other words, you first need to project the vector onto the xy-plane, giving you a vector in [itex]\mathbb{R}^2[/itex] and then find the angle between this projection and the x-axis. Does that make sense?
ibysaiyan said:
Part iv doesn't seem that tricky... expression for acceleration has to be: a = f/m ...
v) Finding out velocity could be find out by using f = momentum /time...
Since the particle is being acted on by a constant force, I would use SUVAT equations.
 
  • #5


HallsofIvy said:
Why in the world would you want to trick your teacher into thinking you know things you don't? You say that "this question comes out of a tutorial sheet where they don't expect us to know much and just see how much knowledge we have retained." If you cannot do this problem yourself, then your teacher needs to know that so he/she can teach it!

Thanks for your reply.I understand where you'e coming from but I am trying to make myself useful here by not wasting around time, I have just found an article describing vectors in good detail.Of what I have found so far... most of my answers on the above post are wrong.
In this case unit vector will be (3.2 4.3 5.4) for part (ii)
So do I need to find out the magnitude of Ox and Oy and then use the above formula to the respective angle ? Am I right ? [for part (iii) ]

But how can I use a formula which is for an angle between two vectors on this... I am confused can someone shed some light on this.. i am not asking for direct answers, a tip maybe ? Thanks
 
  • #6


ibysaiyan said:
But how can I use a formula which is for an angle between two vectors on this... I am confused can someone shed some light on this.. i am not asking for direct answers, a tip maybe ? Thanks
What is the projection of the velocity vector onto the xy-plane?
 
  • #7


Hootenanny said:
What is the projection of the velocity vector onto the xy-plane?

I am not sure rather I have no idea. hm... I have the question posted on the OP.
My current working for all the bits is as following

(i)
speed of particle = square root of [(3.2)^2 + (4.3)^2 + (5.4)^2 ] which gives me an approximate speed of 7.6 m/s

(ii) For unit vector: 3.2 i + 4.3 j + 5.4 k / [7.60 ]

(iii)
theta = tan^-1 ( 4.3/ 3.2) = 53 degree since only axis x and y are mentioned for this part.

(iv) a = f/m

(v) find the velocity at a time of 2s.
Using f = rate of change of momentum / time , we get velocity : force x time / mass
to find force we need to get the modulus of vector force: which is : square root of [ (7.1)^2 + (8.2)^2 + (9.3)^2 ] = ~14.3 N

Velocity at t = 2 ,
14.3 x 2 / 0.1 = 286 m/s.

Please criticize if any of my answer needs to be.
 
  • #8


ibysaiyan said:
I am not sure rather I have no idea. hm... I have the question posted on the OP.
My current working for all the bits is as following

(i)
speed of particle = square root of [(3.2)^2 + (4.3)^2 + (5.4)^2 ] which gives me an approximate speed of 7.6 m/s

(ii) For unit vector: 3.2 i + 4.3 j + 5.4 k / [7.60 ]

(iii)
theta = tan^-1 ( 4.3/ 3.2) = 53 degree since only axis x and y are mentioned for this part.

(iv) a = f/m
All good.
ibysaiyan said:
(v) find the velocity at a time of 2s.
Using f = rate of change of momentum / time , we get velocity : force x time / mass
to find force we need to get the modulus of vector force: which is : square root of [ (7.1)^2 + (8.2)^2 + (9.3)^2 ] = ~14.3 N

Velocity at t = 2 ,
14.3 x 2 / 0.1 = 286 m/s.
Not sure what you're doing here - velocity is a vector, not a scalar.

I suspect that there is some information missing from the question. Could you copy it verbatim please?
 
  • #9


Hootenanny said:
All good.

Not sure what you're doing here - velocity is a vector, not a scalar.

I suspect that there is some information missing from the question. Could you copy it verbatim please?

Sure.

A particle of mass m = 0.1 kg has a velocity vector, v, given by: 3.2 i + 4.3 j + 5.4 k m/s
it's acted by a force
f = 7.1 i + 8.2 j + 9.3 k N

i) Find the speed of the particle
ii) Write down a unit vector in the direction of velocity
iii) Find the angle between the projection of the velocity in the x,y plan and the x-axis.
iv) Find an expression of acceleration.
v) Find the velocity at a time 2 seconds later.


EDIT: Sorry HN I just noticed post no. 4 where you have mentioned suvat equation...
Here's what I have got for the last two parts;
iv) v = u+at => v-u/ t = a.
v) a = f/ m , 14.3 / 0.1 = 143 m/s^2
to find velocity we use the equation stated in part (iv) :
v = u+at = 7.6 + (2 x 143)= ~ 294 m/s.

How about now ?
 
Last edited:
  • #10


ibysaiyan said:
Sure.

A particle of mass m = 0.1 kg has a velocity vector, v, given by: 3.2 i + 4.3 j + 5.4 k m/s
it's acted by a force
f = 7.1 i + 8.2 j + 9.3 k N

i) Find the speed of the particle
ii) Write down a unit vector in the direction of velocity
iii) Find the angle between the projection of the velocity in the x,y plan and the x-axis.
iv) Find an expression of acceleration.
v) Find the velocity at a time 2 seconds later.
Excellent. So, for part (v), since the force is constant and you have the acceleration and initial velocity, I would use SUVAT equations.
 
  • #11


Hootenanny said:
Excellent. So, for part (v), since the force is constant and you have the acceleration and initial velocity, I would use SUVAT equations.

I have posted the solution on the same post. Can you have a look please.Thank you very much for your contribution!
 
  • #12


ibysaiyan said:
I have posted the solution on the same post. Can you have a look please.Thank you very much for your contribution!
ibysaiyan said:
EDIT: Sorry HN I just noticed post no. 4 where you have mentioned suvat equation...
Here's what I have got for the last two parts;
iv) v = u+at => v-u/ t = a.
v) a = f/ m , 14.3 / 0.1 = 143 m/s^2
to find velocity we use the equation stated in part (iv) :
v = u+at = 7.6 + (2 x 143)= ~ 294 m/s.

How about now ?
Again, you have to be careful here. Velocity is a vector quantity, not scalar. You need to apply that SUVAT equation to each component individually. That will then give you the final velocity vector.
 
  • #13


Hootenanny said:
Again, you have to be careful here. Velocity is a vector quantity, not scalar. You need to apply that SUVAT equation to each component individually. That will then give you the final velocity vector.

Hm.. I think I am getting this now...
So I should use suvat equations to find out the initial velocity of the particle and in the same manner for acceleration , which will subsequently get me the final velocity at the time interval of 2 seconds?
 
  • #14


ibysaiyan said:
Hm.. I think I am getting this now...
So I should use suvat equations to find out the initial velocity of the particle and in the same manner for acceleration , which will subsequently get me the final velocity at the time interval of 2 seconds?
No, no. You have been given the initial velocity (v in the question) and you have already worked out the acceleration (a=f/m - remember this is a vector too). Now all you need to do is apply that SUVAT equation to x,y and z components individually to find the final velocity. Do you follow?
 
  • #15


Hootenanny said:
No, no. You have been given the initial velocity (v in the question) and you have already worked out the acceleration (a=f/m - remember this is a vector too). Now all you need to do is apply that SUVAT equation to x,y and z components individually to find the final velocity. Do you follow?

Ah I think I understand the flaw in my reasoning... was it wrong of me to take speed of 7.6m/s into account for my final velocity, since it's a scalar (speed) quantity,right ? but when you speak of components xyz , I presume them to be ijk vector. If this is right then how do I get initial 'v' value for each. Will finding out the magnitude of individual components give me those numbers ?
 
  • #16


ibysaiyan said:
Ah I think I understand the flaw in my reasoning... was it wrong of me to take speed of 7.6m/s into account for my final velocity, since it's a scalar (speed) quantity,right ? but when you speak of components xyz , I presume them to be ijk vectors. If this is right then how do I get initial 'v' value for each. Will finding out the magnitude of individual components give me those numbers ?
I think its best if I show you an example. So, we have the initial velocity and acceleration vectors [itex]\boldsymbol{u} = [3.2, 4.3, 5.4]^\text{T}[/itex] and [itex]\boldsymbol{a} = [71, 82, 93]^\text{T}[/itex]. Now, let's look at the first (x-component), we have: [itex]u_x = 3.2[/itex] and [itex]a_x = 71[/itex]. So, now applying the SUVAT equation

[tex]v = ut + \frac{1}{2}at^2[/tex]

to just the x-component, we find

[tex]v_x = 3.2\times2 + \frac{1}{2}\times 71 \times 2^2 = 148.4 \text{m/s}[/tex]

So, the first component of the final velocity is 148.4 m/s. Now, you need to do the same for the y and z components.

Do you follow?
 
  • #17


Yes, I do now. Thank you very much!
Expect more postage from me. Again I can't thank you enough...
-ibysaiyan
 
  • #18


ibysaiyan said:
Yes, I do now. Thank you very much!
Expect more postage from me. Again I can't thank you enough...
-ibysaiyan
It was a pleasure. Glad to help out.
 

1. What is acceleration?

Acceleration is a measure of the rate at which an object's velocity is changing. It is defined as the change in velocity over a specific period of time.

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What are the units of acceleration?

The units of acceleration depend on the units of velocity and time used in the calculation. However, the standard SI unit for acceleration is meters per second squared (m/s^2).

4. How do you find the expression of acceleration on a velocity-time graph?

The expression of acceleration on a velocity-time graph is the slope of the line. This can be found by calculating the change in velocity over the change in time between two points on the graph.

5. Can acceleration be negative?

Yes, acceleration can be negative. Negative acceleration, also known as deceleration, occurs when an object's velocity decreases over time. This can be seen as a downward slope on a velocity-time graph.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
822
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
14
Views
280
Replies
27
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top