The equation of motion for constant non-zero acceleration a which gives the dependence of the velocity v?
...on the distance travelled is:
(Use x0, v0 for initial position and velocity)
answer should be v^2=
To derive the equation of motion for constant non-zero acceleration, let's start by using the following variables:
- x0: initial position
- v0: initial velocity
- a: constant acceleration
- x: distance traveled (position at any given time t)
- v: velocity at any given time t
Now, we can use the equations of motion to derive the equation for the dependence of velocity on the distance traveled.
1. v = v0 + at (equation of motion relating velocity and time)
To eliminate time from this equation, we need to express it in terms of distance traveled (x). We can do this by utilizing another equation of motion:
2. x - x0 = v0t + (1/2)at^2 (equation of motion relating position and time)
Rearranging equation 2 to solve for time (t):
x - x0 = v0t + (1/2)at^2
(1/2)at^2 + v0t - (x - x0) = 0
This is a quadratic equation in terms of t. We can solve for t using the quadratic formula:
t = (-v0 ± √(v0^2 + 2a(x - x0))) / a
Now, substitute this expression for t into equation 1:
v = v0 + a((-v0 ± √(v0^2 + 2a(x - x0))) / a)
v = v0 - v0 ± √(v0^2 + 2a(x - x0))
v = ± √(v0^2 + 2a(x - x0))
Since we are considering constant non-zero acceleration, the negative sign can be excluded:
v = √(v0^2 + 2a(x - x0))
This equation represents the dependence of velocity (v) on the distance traveled (x), given the initial position (x0), and initial velocity (v0), with a constant non-zero acceleration (a).
Therefore, the final equation is:
v^2 = v0^2 + 2a(x - x0)