Hi so I am currently a 5.2 student at my high school and am wanting to study chemistry next year and was wanting help with a specific set of algebraic questions as my teachers arent helping me much
these are just a few questions so i can get a better understanding of how rearranging works
"
Re-arrange the following equations so that the term indicated is made the subject of the
equation."
Q1.) V= u + at
(i) U
(ii) v= 4, a = 2, t = 1, u = ?
Q2.) v2 = u2 + 2as
(i) u =
(ii) v = 10, a = 4, s = 10, u = ?
Thank you so much in adance to whoever may help me out in my struggles,
Kind Regards,
Callum
Q1.)
(i)
Rearrange v = u + at so that u is made the subject of the equation.
This means: Get u all by itself on one side of the equation.
v = u + at Subtract at from both sides of the equation.
v - at = u + at - at Any number plus fifty bazillion minus fifty bazilion equals the original number.
v - at = u or we can write this as...
u = v - at
(ii)
I think the instructions are...
If u = v - at , what does u equal when v = 4 , a = 2 , and t = 1 ?
u = v - at Replace " v " with 4 , replace " a " with 2 , and replace " t " with 1
u = 4 - (2)(1)
u = 4 - 2
u = 2
Q2.)
(i)
We want to get u all by itself on one side of the equation.
v2 = u2 + 2as Subtract 2as from both sides of the equation.
v2 - 2as = u2 + 2as - 2as
v2 - 2as = u2 Take the square root of both sides of the equation. *
\(\sqrt{v^2-2as}=\sqrt{u^2}\) The square root of any number squared equals the number.
\( \sqrt{v^2-2as}\) = u
u = \( \sqrt{v^2-2as}\)
(ii)
u = \( \sqrt{v^2-2as}\) Replace " v " with 10 , replace " a " with 4 , and replace " s " with 10
u = \( \sqrt{10^2-2(4)(10)}\)
u = \( \sqrt{100-80}\)
u = \( \sqrt{20} \) And the square root of 20 can also be written as...
u = \(2\sqrt5\)
* I only took the positive root because I think that is all you need to do, but I might be wrong. If you want to be a cool kid, you can say u = \( \pm\sqrt{v^2-2as}\) and u = \(\pm2\sqrt5\)
I hope this made some sense! If you have a question don't hesitate to ask!
Q1.)
(i)
Rearrange v = u + at so that u is made the subject of the equation.
This means: Get u all by itself on one side of the equation.
v = u + at Subtract at from both sides of the equation.
v - at = u + at - at Any number plus fifty bazillion minus fifty bazilion equals the original number.
v - at = u or we can write this as...
u = v - at
(ii)
I think the instructions are...
If u = v - at , what does u equal when v = 4 , a = 2 , and t = 1 ?
u = v - at Replace " v " with 4 , replace " a " with 2 , and replace " t " with 1
u = 4 - (2)(1)
u = 4 - 2
u = 2
Q2.)
(i)
We want to get u all by itself on one side of the equation.
v2 = u2 + 2as Subtract 2as from both sides of the equation.
v2 - 2as = u2 + 2as - 2as
v2 - 2as = u2 Take the square root of both sides of the equation. *
\(\sqrt{v^2-2as}=\sqrt{u^2}\) The square root of any number squared equals the number.
\( \sqrt{v^2-2as}\) = u
u = \( \sqrt{v^2-2as}\)
(ii)
u = \( \sqrt{v^2-2as}\) Replace " v " with 10 , replace " a " with 4 , and replace " s " with 10
u = \( \sqrt{10^2-2(4)(10)}\)
u = \( \sqrt{100-80}\)
u = \( \sqrt{20} \) And the square root of 20 can also be written as...
u = \(2\sqrt5\)
* I only took the positive root because I think that is all you need to do, but I might be wrong. If you want to be a cool kid, you can say u = \( \pm\sqrt{v^2-2as}\) and u = \(\pm2\sqrt5\)
I hope this made some sense! If you have a question don't hesitate to ask!