- #1
LoveBoy
- 44
- 1
When we apply this equation ?
v^2-u^2=kt^2/2m
v^2-u^2=kt^2/2m
Not sure what that equation is supposed to be. It looks like a mash up of these two:LoveBoy said:When we apply this equation ?
v^2-u^2=kt^2/2m
Thanks !Doc Al said:Not sure what that equation is supposed to be. It looks like a mash up of these two:
##x = x_0 + v_0 t + (1/2) a t^2##
##v^2 = v_0^2 + 2 a \Delta x##
To solve for time in a kinematics equation, you need to have values for the other variables in the equation, such as displacement, velocity, and acceleration. Once you have these values, you can use the appropriate kinematics equation to solve for time.
There are three main kinematics equations that can be used to solve for time: t = (vf - vi)/a, t = d/v, and t = 2d/(v + vf). These equations depend on the specific problem and the given values. You can also use the quadratic formula in some cases.
In most cases, you need to have values for at least two other variables in order to solve for time using a kinematics equation. However, if you have a constant acceleration and know the displacement, you can use the equation t = √(2d/a) to solve for time.
One common mistake is not using the correct equation for the given problem. It's important to understand which equation to use based on the given variables. Another mistake is not paying attention to units - make sure all values are in the correct units before plugging them into the equation.
Kinematics equations can be used to solve for time in most situations involving constant acceleration. However, they may not be applicable in cases of non-constant acceleration, such as when an object is accelerating due to a changing force or when there is a combination of acceleration and deceleration. In these cases, calculus may be needed to find the time variable.