- #1

- 1,113

- 125

- Homework Statement
- Please see below

- Relevant Equations
- Please see below

For ## d = v_it + 0.5at^2##, can we solve for t without using the quadratic formula?

Many thanks!

Many thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 1,113

- 125

- Homework Statement
- Please see below

- Relevant Equations
- Please see below

For ## d = v_it + 0.5at^2##, can we solve for t without using the quadratic formula?

Many thanks!

Many thanks!

- #2

- 24,226

- 15,960

You can solve a quadratic equation by *completing* *the* *square*.

- #3

- 1,113

- 125

Oh true, thank you @PeroK !You can solve a quadratic equation bycompletingthesquare.

- #4

- 39,852

- 9,043

Which is equivalent to using the formula, no?You can solve a quadratic equation bycompletingthesquare.

- #5

- 24,226

- 15,960

No. If you forget the formula you can still complete the square. Or, vice versa!Which is equivalent to using the formula, no?

- #6

- 279

- 175

The formula is derived by completing the square for the general case of a 2nd degree polynomial. Regardless, completing the square is a technique that is good to know as you will sometimes have to rewrite expressions "in-place", i.e. by subtracting "zero" and what not rather than moving terms from LHS->RHS.Which is equivalent to using the formula, no?

- #7

Homework Helper

- 1,017

- 535

$${v_f}^2={v_i}^2+2ad$$Homework Statement::Please see below

Relevant Equations::Please see below

For ## d = v_it + 0.5at^2##, can we solve for t without using the quadratic formula?

Many thanks!

$$t=\frac{v_f-v_i}{a}$$

- #8

- 39,852

- 9,043

Again, not really any different from using the formula.$${v_f}^2={v_i}^2+2ad$$

$$t=\frac{v_f-v_i}{a}$$

@Callumnc1 , why do you wish to avoid using the formula?

- #9

Homework Helper

- 4,179

- 765

Homework Statement::Please see below

Relevant Equations::Please see below

For ## d = v_it + 0.5at^2##, can we solve for t without using the quadratic formula?

Many thanks!

So, you are asking if you can solve a quadratic equation without using the quadratic formula? Or it is something else?

- #10

- 197

- 74

I think it should have same result but the suggested method helps us find roots withhout knowing how to solve a quadratic equation. He finds them using constant acceleration equations ...Again, not really any different from using the formula.

@Callumnc1 , why do you wish to avoid using the formula?

- #11

- 1,113

- 125

Thank you for your reply @haruspex !Which is equivalent to using the formula, no?

- #12

- 1,113

- 125

Thank you for your reply @PeroK!No. If you forget the formula you can still complete the square. Or, vice versa!

- #13

- 1,113

- 125

Thank you for your reply @Mayhem!The formula is derived by completing the square for the general case of a 2nd degree polynomial. Regardless, completing the square is a technique that is good to know as you will sometimes have to rewrite expressions "in-place", i.e. by subtracting "zero" and what not rather than moving terms from LHS->RHS.

- #14

- 1,113

- 125

Thank you for your reply @neilparker62 !$${v_f}^2={v_i}^2+2ad$$

$$t=\frac{v_f-v_i}{a}$$

- #15

- 1,113

- 125

Thank you for your reply @haruspex !Again, not really any different from using the formula.

@Callumnc1 , why do you wish to avoid using the formula?

I was just curious :)

- #16

- 1,113

- 125

Thank you for your reply @nasu! Yeah, I guess you think what I'm asking like that!So, you are asking if you can solve a quadratic equation without using the quadratic formula? Or it is something else?

- #17

- 1,113

- 125

Thank you for your reply @MatinSAR!I think it should have same result but the suggested method helps us find roots withhout knowing how to solve a quadratic equation. He finds them using constant acceleration equations ...

- #18

Homework Helper

- 1,017

- 535

$$d=v_it+½ at^2$$ $$\frac{a}{2} \left(t^2 + \frac{2v_it}{a}-\frac{2d}{a}\right) = 0 $$ $$\frac{a}{2} \left[ \left(t+\frac{v_i}{a}\right)^2-\frac{2ad}{a^2} - \left(\frac{v_i}{a}\right)^2\right] = 0 $$ $$t=\frac{\sqrt{{v_i}^2+2ad} - v_i}{a}=\frac{v_f-v_i}{a}$$

- #19

- 197

- 74

$$d=v_it+½ at^2$$ $$\frac{a}{2} \left(t^2 + \frac{2v_it}{a}-\frac{2d}{a}\right) = 0 $$ $$\frac{a}{2} \left[ \left(t+\frac{v_i}{a}\right)^2-\frac{2ad}{a^2} - \left(\frac{v_i}{a}\right)^2\right] = 0 $$ $$t=\frac{\sqrt{{v_i}^2+2ad} - v_i}{a}=\frac{v_f-v_i}{a}$$

Using constant acceleration equations you have mentioned in post #7 we can find both roots without formula:

##d=v_it+\frac 1 2 at^2## ##\rightarrow## ##v_f=v_i+at## ##\rightarrow## ##t=\frac {v_f-v_i}{a}##

##v_f=±\sqrt {v_i^2+2ad}##

##t=\frac {±\sqrt {v_i^2+2ad}-v_i}{a}##

- #20

- 1,113

- 125

Thank you @neilparker62 for showing me the completing square method! I was thinking how I would do that!

$$d=v_it+½ at^2$$ $$\frac{a}{2} \left(t^2 + \frac{2v_it}{a}-\frac{2d}{a}\right) = 0 $$ $$\frac{a}{2} \left[ \left(t+\frac{v_i}{a}\right)^2-\frac{2ad}{a^2} - \left(\frac{v_i}{a}\right)^2\right] = 0 $$ $$t=\frac{\sqrt{{v_i}^2+2ad} - v_i}{a}=\frac{v_f-v_i}{a}$$

- #21

- 1,113

- 125

Thank you @MatinSAR for showing me your method!Using constant acceleration equations you have mentioned in post #7 we can find both roots without formula:

##d=v_it+\frac 1 2 at^2## ##\rightarrow## ##v_f=v_i+at## ##\rightarrow## ##t=\frac {v_f-v_i}{a}##

##v_f=±\sqrt {v_i^2+2ad}##

##t=\frac {±\sqrt {v_i^2+2ad}-v_i}{a}##

- #22

Homework Helper

- 1,017

- 535

Yes - I should have included the ##\pm##. Then you might call it the kinematics version of the quadratic formula! Good idea to reflect upon when you would use "+" and when "-" in the formula. Also consider the physical meaning of the quadratic's axis of symmetry and its maximum value in the context of projectile motion: $$h(t) = v_it - ½gt^2=\frac{-g}{2} \left[ \left(t-\frac{v_i}{g}\right)^2 - \left(\frac{v_i}{-g}\right)^2\right].$$ Assume up is the positive direction and g is the standard acceleration of gravity: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.Using constant acceleration equations you have mentioned in post #7 we can find both roots without formula:

##d=v_it+\frac 1 2 at^2## ##\rightarrow## ##v_f=v_i+at## ##\rightarrow## ##t=\frac {v_f-v_i}{a}##

##v_f=±\sqrt {v_i^2+2ad}##

##t=\frac {±\sqrt {v_i^2+2ad}-v_i}{a}##

Last edited:

- #23

- 1,113

- 125

Thank you for your reply @neilparker62 !Yes - I should have included the ##\pm##. Then you might call it the kinematics version of the quadratic formula! Good idea to reflect upon when you would use "+" and when "-" in the formula. Also consider the physical meaning of the quadratic's axis of symmetry and its maximum value in the context of projectile motion: $$h(t) = v_it - ½gt^2=\frac{-g}{2} \left[ \left(t-\frac{v_i}{g}\right)^2 - \left(\frac{v_i}{-g}\right)^2\right].$$ Assume up is the positive direction and g is the standard acceleration of gravity: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.

Sorry, could please explain how you got from ##h(t) = v_it - ½gt^2## to ##h(t) = \frac{-g}{2} \left[ \left(t-\frac{v_i}{g}\right)^2 - \left(\frac{v_i}{-g}\right)^2\right]##?

Many thanks!

- #24

Homework Helper

- 1,017

- 535

Largely the same process as in post #18. I just wrote d as "h(t)" and did not move it onto the right hand side. Otherwise followed standard procedure for completing the square (remove coefficient of t^2 and add/subtract "half the coefficient of t" ^2.)

In Latex, I simply copied the line from post #18 and edited it to remove the term with d in it and replace "a" with "-g" everywhere.

In Latex, I simply copied the line from post #18 and edited it to remove the term with d in it and replace "a" with "-g" everywhere.

Last edited:

- #25

- 1,113

- 125

Ok thank you very much for your help @neilparker62 !Largely the same process as in post #18. I just wrote d as "h(t)" and did not move it onto the right hand side. Otherwise followed standard procedure for completing the square (remove co-efficient of t^2 and add/subtract "half the coefficient of t" ^2.)

In Latex, I simply copied the line from post #18 and edited it to remove the term with d in it and replace "a" with "-g" everywhere.

Share:

- Replies
- 25

- Views
- 1K

- Replies
- 11

- Views
- 384

- Replies
- 2

- Views
- 467

- Replies
- 22

- Views
- 432

- Replies
- 8

- Views
- 382

- Replies
- 16

- Views
- 1K

- Replies
- 6

- Views
- 398

- Replies
- 1

- Views
- 507

- Replies
- 8

- Views
- 695

- Replies
- 8

- Views
- 283