# Solving kinematic formula for t

• ChiralSuperfields
In summary: 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}####v_f=±\sqrt {v_i^2+2ad}####t=\frac {±\sqrt {v_i^2+2ad}-v_i}{a}
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
PeroK said:
You can solve a quadratic equation by completing the square.
Oh true, thank you @PeroK !

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

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

ChiralSuperfields
haruspex said:
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
Callumnc1 said:

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
neilparker62 said:
$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$
Again, not really any different from using the formula.
@Callumnc1 , why do you wish to avoid using the formula?

ChiralSuperfields and MatinSAR
Callumnc1 said:

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
haruspex said:
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
haruspex said:
Which is equivalent to using the formula, no?

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

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

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

I was just curious :)

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

MatinSAR said:
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
neilparker62 said:
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
neilparker62 said:
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!

MatinSAR said:
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
MatinSAR said:
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
neilparker62 said:
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.

Last edited:
ChiralSuperfields
neilparker62 said:
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 !

## 1. How do I solve kinematic formulas for time (t)?

To solve kinematic formulas for time, you will need to have three out of the four variables: displacement (d), initial velocity (v0), final velocity (v), and acceleration (a). You can then use the appropriate kinematic formula and algebraic manipulation to solve for time (t).

## 2. What is the formula for solving for time in kinematics?

The formula for solving for time in kinematics depends on the given variables. Some common kinematic formulas for time include t = (v - v0) / a, t = d / v0, and t = √(2d / a).

## 3. Can I use kinematic formulas to solve for time in any situation?

Kinematic formulas can only be used to solve for time in situations where the acceleration is constant. If the acceleration is not constant, calculus or other methods may be necessary to solve for time.

## 4. How do I know which kinematic formula to use to solve for time?

The kinematic formula to use depends on the given variables and what you are trying to solve for. It is important to carefully read the problem and identify which variables are known and which one you are trying to solve for. Then, choose the appropriate formula and plug in the values.

## 5. What are some common mistakes to avoid when solving kinematic formulas for time?

Some common mistakes to avoid when solving kinematic formulas for time include using the wrong formula, not converting units properly, and not paying attention to the signs of the variables. It is also important to check your final answer to make sure it is reasonable and makes sense in the context of the problem.

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