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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

Guest Jun 5, 2017

Best Answer 

 #1
avatar+4694 
+2

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  =  \(\pm2\sqrt5\)     winklaugh

 

 

I hope this made some sense! If you have a question don't hesitate to ask!  smiley

hectictar  Jun 5, 2017
Sort: 

1+0 Answers

 #1
avatar+4694 
+2
Best Answer

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  =  \(\pm2\sqrt5\)     winklaugh

 

 

I hope this made some sense! If you have a question don't hesitate to ask!  smiley

hectictar  Jun 5, 2017

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