This is a cubic equation in v. There is a formula for calculating the roots of a cubic equation, but it is much more complicated than the quadratic formula for sloving quadratic equations.MeL82vin said:Homework Statement
423,000=5504v+4.37v^3
2. Questions
Could someone please advise on how to solve for velocity(v)? Our tutor solved the equation using equation solver software to get 36.9 m/s (133km/h) but he included it in our exam and wanted a manual calculation of the velocity... It's beyond me, I am just trying to gauge how difficult it is to obtain the answer as I find it unusual that he had to use software to get a value but expected us to calculate it manually.
Now that you have mentioned "trial and error" I'm fairly sure that's the method he wanted us to use. It didn't seem so obvious at the time though! Hopefully it doesn't catch me out again, thanks for the help.SteamKing said:This is a cubic equation in v. There is a formula for calculating the roots of a cubic equation, but it is much more complicated than the quadratic formula for sloving quadratic equations.
https://en.wikipedia.org/wiki/Cubic_function
Rather than memorizing several pages of formulas, it's probably easier to use a trial and error method to find a root.
To solve for velocity in this equation, we need to isolate the "v" term on one side of the equation. We can do this by subtracting 5504v from both sides, giving us 417,496 = 4.37v^3. Then, we can divide both sides by 4.37 to get v^3 = 95,552. Finally, we can take the cube root of both sides to find that v = 44.07 m/s.
In this equation, the units for velocity are meters per second (m/s). This is because the units for the constant terms (423,000 and 5504) are both in meters, and the units for the variable term (v^3) are in meters cubed. Therefore, we can solve for velocity in m/s to maintain consistency in units.
Yes, this equation can be solved with other units for velocity, as long as the units are consistent on both sides of the equation. For example, if we wanted to solve for velocity in kilometers per hour (km/h), we would need to convert the constant terms to kilometers and the variable term to kilometers cubed before solving for velocity.
This equation can be used in a variety of real-world situations where we know the values for the other terms in the equation. For example, if we know the acceleration and displacement of an object, we can use the equation v^2 = u^2 + 2as to solve for the velocity. Then, we can substitute this value for velocity into the original equation to solve for the constant terms, which can give us a better understanding of the object's motion.
Yes, there are some limitations to using this equation to solve for velocity. One limitation is that it assumes uniform acceleration, which may not always be the case in real-world scenarios. Additionally, this equation only works for one-dimensional motion and does not take into account other factors such as air resistance or friction. Therefore, it should be used with caution and in conjunction with other equations and factors when solving for velocity in real-world situations.