- #1
ChiralSuperfields
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- Homework Statement
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- Relevant Equations
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For ## d = v_it + 0.5at^2##, can we solve for t without using the quadratic formula?
Many thanks!
Many thanks!
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?PeroK said:You can solve a quadratic equation by completing the square.
No. If you forget the formula you can still complete the square. Or, vice versa!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.haruspex said:Which is equivalent to using the formula, no?
$${v_f}^2={v_i}^2+2ad$$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!
Again, not really any different from using the formula.neilparker62 said:$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$
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!
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 ...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 !haruspex said:Which is equivalent to using the formula, no?
Thank you for your reply @PeroK!PeroK said:No. If you forget the formula you can still complete the square. Or, vice versa!
Thank you for your reply @Mayhem!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 @neilparker62 !neilparker62 said:$${v_f}^2={v_i}^2+2ad$$
$$t=\frac{v_f-v_i}{a}$$
Thank you for your reply @haruspex !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 @nasu! Yeah, I guess you think what I'm asking like that!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 @MatinSAR!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 ...
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!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 @MatinSAR for showing me your method!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: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.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 for your reply @neilparker62 !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: $$g = 9.806 65 ms^{-2} \approx 9.8 ms^{-2}$$.
Ok thank you very much for your help @neilparker62 !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.
To solve kinematic formulas for time, you will need to have three out of the four variables: displacement (d), initial velocity (v_{0}), final velocity (v), and acceleration (a). You can then use the appropriate kinematic formula and algebraic manipulation to solve for time (t).
The formula for solving for time in kinematics depends on the given variables. Some common kinematic formulas for time include t = (v - v_{0}) / a, t = d / v_{0}, and t = √(2d / a).
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.
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.
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.