# Solving kinematic formula for t

• ChiralSuperfields

#### ChiralSuperfields

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

Many thanks!

You can solve a quadratic equation by completing the square.

• MatinSAR and ChiralSuperfields
You can solve a quadratic equation by completing the square.
Oh true, thank you @PeroK !

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

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

• ChiralSuperfields
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.

• ChiralSuperfields and PeroK

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

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

• ChiralSuperfields and MatinSAR
$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$
Again, not really any different from using the formula.
@Callumnc1 , why do you wish to avoid using the formula?

• ChiralSuperfields and MatinSAR

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?

• ChiralSuperfields
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 and SammyS
Which is equivalent to using the formula, no?

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

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.

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

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

I was just curious :)

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

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 ...

• MatinSAR
Completing the square:

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

• ChiralSuperfields, PeroK and 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}##

• neilparker62, ChiralSuperfields and PeroK
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!

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}##

Thank you @MatinSAR for showing me your method!

• MatinSAR
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}##

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

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• ChiralSuperfields and MatinSAR
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!

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.

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• 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 !