Solving kinematic formula for t

[ChiralSuperfields]

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!

 

[PeroK]

You can solve a quadratic equation by completing the square.

 

[ChiralSuperfields]

You can solve a quadratic equation by completing the square.

Oh true, thank you @PeroK !

 

[haruspex]

You can solve a quadratic equation by completing the square.

Which is equivalent to using the formula, no?

 

[PeroK]

Which is equivalent to using the formula, no?

No. If you forget the formula you can still complete the square. Or, vice versa!

 

[Mayhem]

Which is equivalent to using the formula, no?

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.

 

[neilparker62]

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!

$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$

 

[haruspex]

$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$

Again, not really any different from using the formula.
@Callumnc1 , why do you wish to avoid using the formula?

 

[nasu]

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?

 

[MatinSAR]

Again, not really any different from using the formula.
@Callumnc1 , why do you wish to avoid using the formula?

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 ...

 

[ChiralSuperfields]

Which is equivalent to using the formula, no?

Thank you for your reply @haruspex !

 

[ChiralSuperfields]

No. If you forget the formula you can still complete the square. Or, vice versa!

Thank you for your reply @PeroK!

 

[ChiralSuperfields]

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.

Thank you for your reply @Mayhem!

 

[ChiralSuperfields]

$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$

Thank you for your reply @neilparker62 !

 

[ChiralSuperfields]

Again, not really any different from using the formula.
@Callumnc1 , why do you wish to avoid using the formula?

Thank you for your reply @haruspex !

I was just curious :)

 

[ChiralSuperfields]

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

Thank you for your reply @nasu! Yeah, I guess you think what I'm asking like that!

 

[ChiralSuperfields]

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 ...

Thank you for your reply @MatinSAR!

 

[neilparker62]

Completing the square:

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

 

[MatinSAR]

Completing the square:

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

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}##

 

[ChiralSuperfields]

Completing the square:

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

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

 

[ChiralSuperfields]

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}##

Thank you @MatinSAR for showing me your method!

 

[neilparker62]

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}##

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: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.

 

[ChiralSuperfields]

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: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.

Thank you for your reply @neilparker62 !

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!

 

[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 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.

 

[ChiralSuperfields]

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.

Ok thank you very much for your help @neilparker62 !