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, we can use the kinematic equation. The equation relates the final velocity (v) of an object to its initial velocity (v0), acceleration (a), and distance traveled (x).
Here's how you can derive the equation:
1. Start with the equation of motion relating displacement (x), initial velocity (v0), final velocity (v), and acceleration (a):
x = v0t + (1/2)at^2, where t represents time.
2. Let's eliminate time from the equation. We can solve the equation for t using the equation of motion for final velocity and acceleration:
v = v0 + at
Rearranging the equation, we get:
t = (v - v0) / a
3. Substitute the expression for time (t) into the displacement equation:
x = v0((v - v0) / a) + (1/2)a((v - v0) / a)^2
4. Simplify the equation by canceling out common factors:
x = (v(v - v0) / a) + (1/2)(v - v0)^2 / a
5. Multiply through by a to remove the denominator:
ax = v(v - v0) + (1/2)(v - v0)^2
6. Distribute and simplify:
ax = v^2 - v0v + (1/2)(v^2 - 2v0v + v0^2)
7. Combine like terms:
ax = v^2 - v0v + (1/2)v^2 - v0v + (1/2)v0^2
8. Simplify further:
ax = (3/2)v^2 - 2v0v + (1/2)v0^2
9. Rearrange the equation:
(3/2)v^2 = ax + 2v0v - (1/2)v0^2
10. Finally, divide both sides by (3/2) to isolate v^2:
v^2 = (2/3)(ax + 2v0v - (1/2)v0^2)
Therefore, the equation of motion for constant non-zero acceleration in terms of velocity squared is:
v^2 = (2/3)(ax + 2v0v - (1/2)v0^2)
Note: In the given question, it mentions the dependence of velocity on the distance traveled, which suggests a different equation. However, based on the given variables (x0, v0), we can derive the equation of motion in terms of velocity squared as shown above.