Solving kinematic formula for t

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}
  • #1
ChiralSuperfields
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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!
 
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  • #2
You can solve a quadratic equation by completing the square.
 
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  • #3
PeroK said:
You can solve a quadratic equation by completing the square.
Oh true, thank you @PeroK !
 
  • #4
PeroK said:
You can solve a quadratic equation by completing the square.
Which is equivalent to using the formula, no?
 
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  • #5
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!
 
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  • #6
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.
 
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  • #7
Callumnc1 said:
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!
$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$
 
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  • #8
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?
 
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  • #9
Callumnc1 said:
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?
 
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  • #10
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 ...
 
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  • #11
haruspex said:
Which is equivalent to using the formula, no?
Thank you for your reply @haruspex !
 
  • #12
PeroK said:
No. If you forget the formula you can still complete the square. Or, vice versa!
Thank you for your reply @PeroK!
 
  • #13
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.
Thank you for your reply @Mayhem!
 
  • #14
neilparker62 said:
$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$
Thank you for your reply @neilparker62 !
 
  • #15
haruspex said:
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 :)
 
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  • #16
nasu said:
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!
 
  • #17
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 ...
Thank you for your reply @MatinSAR!
 
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  • #18
Completing the square:

$$d=v_it+½ at^2$$ $$\frac{a}{2} \left(t^2 + \frac{2v_it}{a}-\frac{2d}{a}\right) = 0 $$ $$\frac{a}{2} \left[ \left(t+\frac{v_i}{a}\right)^2-\frac{2ad}{a^2} - \left(\frac{v_i}{a}\right)^2\right] = 0 $$ $$t=\frac{\sqrt{{v_i}^2+2ad} - v_i}{a}=\frac{v_f-v_i}{a}$$
 
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  • #19
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}##
##v_f=±\sqrt {v_i^2+2ad}##

##t=\frac {±\sqrt {v_i^2+2ad}-v_i}{a}##
 
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  • #20
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!
 
  • #21
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}##
##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!
 
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  • #22
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}##
##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: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.
 
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  • #23
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: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.
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!
 
  • #24
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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  • #25
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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