Find average acceleration - vectors

[Michelle027]

1. A car is traveling 9m/s Northwest. 8 seconds later it has rounded a corner and is now heading North at 15m/s.

This was a question from my textbook and was an example question - so they supplied answers. I was able to work through all of it finding everything except the last question.
They wanted me to calculate average acceleration. ( the average acceleration vector was 0.795m/s^2 xhat and 1.08m/s^2 yhat. That I was able to calculate)

Homework Equations

I expected the normal acceleration would have been to use v=vo +at where I interchange the equation to make a the formula of the equation and work out acceleration accordingly. [/B]

The Attempt at a Solution

My attempt was as follows: 24= 0 + a(8) which follows to a= 24/8= 6m/s^2. Which is SO far out from their solution which was 1.34m/s^2[/B]

 

[Orodruin]

It is unclear what they mean by "average acceleration" unless they refer to "magnitude of average acceleration vector". If they mean anything else, such as the average of the magnitude of the acceleration vector, there is too little information.

 

[gneill]

Hi Michelle027, Welcome to Physics Forums!

I suspect that by "normal acceleration" they may be looking for the magnitude of the average acceleration vector that you already found, that is, the magnitude of $(0.795 \hat{x} + 1.08 \hat{y})m/s^2$

Edit: Ah. I see that Orodruin got there before me!

 

[PeroK]

1. A car is traveling 9m/s Northwest. 8 seconds later it has rounded a corner and is now heading North at 15m/s.

This was a question from my textbook and was an example question - so they supplied answers. I was able to work through all of it finding everything except the last question.
They wanted me to calculate average acceleration. ( the average acceleration vector was 0.795m/s^2 xhat and 1.08m/s^2 yhat. That I was able to calculate)

Homework Equations

I expected the normal acceleration would have been to use v=vo +at where I interchange the equation to make a the formula of the equation and work out acceleration accordingly. [/B]

The Attempt at a Solution

My attempt was as follows: 24= 0 + a(8) which follows to a= 24/8= 6m/s^2. Which is SO far out from their solution which was 1.34m/s^2[/B]

Judging from the answer, they mean that the car took 8 seconds to round the corner, turning in some sort of curve.

You don't have constant acceleration in this case. Do you know why not?

In this case, where acceleration is changing, what is the definition of average acceleration?

 

[Michelle027]

Judging from the answer, they mean that the car took 8 seconds to round the corner, turning in some sort of curve.

You don't have constant acceleration in this case. Do you know why not?

In this case, where acceleration is changing, what is the definition of average acceleration?

Thank you for answering
I know that a change in direction brings about acceleration. I presume that is why the acceleration is not constant? I will try a different equation.

 

[Orodruin]

You don't have constant acceleration in this case. Do you know why not?

This is not necessarily true. A constant acceleration would also result in "some sort of curve".

 

[Michelle027]

Hi Michelle027, Welcome to Physics Forums!

I suspect that by "normal acceleration" they may be looking for the magnitude of the average acceleration vector that you already found, that is, the magnitude of $(0.795 \hat{x} + 1.08 \hat{y})m/s^2$

Edit: Ah. I see that Orodruin got there before me!

Thank you so much for replying. I will try that and then make sure I understand why I had to work with that magnitude. I am studying through a distance-education institution so I rely on Google and Youtube and this forum

 

[Michelle027]

This is not necessarily true. A constant acceleration would also result in "some sort of curve".

Oh my word - I will read through that chapter again - doing this through a distance learning institution and it seems the more I try to figure this the more distance there comes between me and understanding how this works! Thank you for your time and replies

 

[PeroK]

Thank you for answering
I know that a change in direction brings about acceleration. I presume that is why the acceleration is not constant? I will try a different equation.

You seem to have worked all this out already, but not in your attempted solution.

If you calculate the magnitude of your average acceleration vector what do you get?

 

[Michelle027]

You seem to have worked all this out already, but not in your attempted solution.

If you calculate the magnitude of your average acceleration vector what do you get?

The average acceleration vector was 0.795m/s^2 xhat and 1.08m/s^2 yhat. That I was able to calculate.

 

[Orodruin]

The average acceleration vector was 0.795m/s^2 xhat and 1.08m/s^2 yhat. That I was able to calculate.

So what is the magnitude of this vector?

 

[Michelle027]

I got initial velocity vector 9m by calculating 9cos 45 ( as direction was NE) that gave me 6.36 as x-axis is negative for the direction given I got xhat -6.36 and yhat will then be 6.36 both in m/s.

Final velocity vectors was yhat 15m/s as x is 0 for that direction which was North.

From that I calculated change in velocity vectors as xhat 6.36 m/s plus 8.64 m/s ( 15 minus 6.36)
Dividing both of those velocity vectors by 8 seconds I came to:
Xhat 0.795 m/s^2 + yhat 1.08 m/s^2
(They did ask for theta and I got that as 53.6 degrees just as the example did. )

 

[PeroK]

Can you see a relationship between the vector $(0.795, 1.08)$ and the scalar $1.34$?

 

[Michelle027]

I Have been trying to figure out what they did to get that I even tried to use the original distances - but Iam still trying ... Maybe I am just missing something.

 

[PeroK]

I Have been trying to figure out what they did to get that I even tried to use the original distances - but Iam still trying ... Maybe I am just missing something.

Your last hint is what do you think good old Pythagoras would say?

 

[Orodruin]

If I give you the vector $2\hat x + 3\hat y$, what is the length of that vector?

 

[Michelle027]

Your last hint is what do you think good old Pythagoras would say?

I thank thee with my hat in my hand... I am just apalled that I didnt try Pythagoras... which tells me I do not fully understand this work so I will go and STUDY SOME MORE...

 

[Michelle027]

If I give you the vector $2\hat x + 3\hat y$, what is the length of that vector?

This is a Pythagoras one, right? I will use the sum of the squares of 2 and 3 and the square root of the answer of those which will be length of the hypotenuse and thus the length of the vector...

 

[Orodruin]

This is a Pythagoras one, right? I will use the sum of the squares of 2 and 3 and the square root of the answer of those which will be length of the hypotenuse and thus the length of the vector...

Right. So you can do the same to your problem.

 

[Michelle027]

Right. So you can do the same to your problem.

Thank you I appreciate