All Answers 358087 Answers

Thanks Alan 

Not clear to me either.  The arrangement of the two objects isn't specified (Is one on top of the other?  If so, which is which?); what does the 2-4 represent (no units are given); where are the forces applied?

. 

Thanks Alan

I understood 3.3, it was the other 2 I was questioning.    So I look forward to your extended explanation.  

I don't know.  He/she might want the total derivative in the form dF = f_xdx + f_ydy + f_zdz, but, if so, the reply in the other thread should help.

.

Q3.3   For an object to be accelerating it must either be changing speed or direction or both.  The question states the object is moving at constant speed, so it can only be accelerating when it is changing direction.  At point C it is not changing direction, it is travelling in a straight line.  So, at C it is changing neither speed nor direction; therefore it is a point of zero acceleration.  At the other points its direction is changing, hence it is accelerating, hence they are not points of zero acceleration.

Q3.7 and 3.8 I'll do a more complete analysis with a diagram later. 

So anonymous just want the three partial derivatives then Alan?

Thanks Alan  

Do it on the site calc on the home page.

(5/6)/(1/6)=

Or you can jusrt invert the second one and change the divide to a multiply

5/6*6/1=30/6=5

It is a notation that is sometimes used, especially in more advanced work.  Not usually at secondary school level though.

.

xvxvxv wrote: how when x-x0 = 150   , t=5s ?  I think t must equal to 10 s ? 

The total time from the start when the ball was at rest is indeed 10 seconds.  However, the problem is solved in such a way that time has been reset to zero for the second part of the problem.  I can understand how you might have been misled, because the question doesn't reset the start of the distance to zero - it calls the start of the second part x₀.  If you like you could solve the question using t set at 10 seconds, but then you must use the equation in the form:  x-x₀ = v_0x(t-t₀)+(1/2)a_x(t-t₀)², where t₀ is 5 seconds.

.

@@ End of Day Wrap            Tues 14/10/14 Sydney,            Sydney,  Australia Time 6:20pm     ♬

Hi all,

Today our fabulous answereres were  Geno3141, CPhill, Alan, Baysour8989, Unknown, Naomi390323, gliu1, Bugzy28871, and Heureka.  Thank you all.   

I didn't notice a lot of interesting posts.  I probably need answerers to point them out more often.

But here are a couple.  I found the first one interesting although I have had no feed back on my answer yet.

It is good when askers are persistent in seeking an answer that they can fully understand.

1) Algebra concepts   Thanks Geno, Chris and Melody.

Wed 14/10/14

1) a bit of light fun.  

Can you word your question a little more clearly please?

Thanks Alan,

I wasn't supposed to recognise this notation was I?

I assume f_y and f_z are partial derivatives with respect to y and z respectively.  If so, then

f = xy + ln(yz)

f_y = x +1/y

f_z = 1/z

.

At the beginning of the second 5 second period the velocity is 5a    (so u=5a)

We are looking at the next 5 seconds so t=5

the distance travelled during this 5 seconds is 150 m

Using the formula

$$\\s=ut+\frac{1}{2}at^2\\\\
150=5a\times 5+\frac{1}{2}a\times5^2\\\\
150=25a+\frac{25a}{2}\\\\
300=50a+25a\\\\
300=75a\\\\
a=4\\\\$$

I've gotten a bit carried away here - have I help you with your dilemma?

Alan,   may I ask a question too?

In Q3.7   What if the i acceleration was huge, would it still have to be slowing down since the j accel is negative?

Isn't the i component of any consequnce at all to slowing down or speeding up if the the original velocity in the i direction is 0?

I'm interested in Q3.8 as well.   (there is an extra ^2 in the i velocity component that shouldn't be there)

It is curved

It is getting faster in the i direction and slower in the j direction.

How do you work more than that out?

I saw this question before (the first time you posted) but it doesn't make sense to me.  Sorry.

2. Find the ordered pairs that are both solutions of y - 4x = 0.

There are an infinite number of ordered pair solutions.

hers are some

(1,4) (2,8)  ....( -6,-24)

to get (1,4)

i just randomly chose 1

then I put 1 in for x

y-4*1=0

y-4=0

y=4

I did the same with the others.     

$$\sum_{n=0}^{\infty} 8.5n=\infty$$

However 

8.5+2*8.5+3*8.5.......n*8.5

is the sum of a geometric series

a=8.5

r=8.5

$$\\s_n=\frac{a(r^n-1)}{r-1}\\\\
s_n=\frac{8.5(8.5^n-1)}{8.5-1}\\\\
s_n=\frac{8.5(8.5^n-1)}{7.5}\\\\$$

I think I got the formula right. 

$$\\4x^2+2x+5=0\\
$There are no integer zeros for this so$\\
$solve it using the quadratic formula.$$$

2x+5y=3

if x=-3

-6+5y=3

5y=9

y=9/5

use the calc on the home page.

Your question doesn't make sense.