Melody

Can you word your question a little more clearly please?

Thanks Alan,

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

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. 

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

use the calc on the home page.

Your question doesn't make sense.