1. One-Dimensional Motion
Physics 1

2. Constant Velocity

3. Constant Velocity
The xd-t graph for constant velocity is linear. A common equation for any line is y = mx + b. In the graph, m is velocity (v), b is initial position (xi), and y is the final position (xf) after a time t.

4. Constant Velocity By substitution,
Subtracting xi to the left hand side,
Since xf – xi = Dx, this results in the expression
Eq. 1

5. Constant Acceleration

6. Constant Acceleration
The v-t graph for constant acceleration is linear. A common equation for any line is y = mx + b. In the graph, m is acceleration (a), b is initial velocity (vi), and y is the final velocity (vf) after a time t. By substitution,
Eq. 2

7. Constant Acceleration
To find the displacement (Dx), determine the area under the v-t graph. The area can be broken into a rectangle and a triangle. The rectangle’s area is bh, where b is t and h is vi. The triangle’s are is ½bh, where b is t and h is (vf – vi).

8. Constant Acceleration
The displacement is equal to the area of the rectangle and the area of the triangle.
Dx = area of █ + area of ▲

9. Constant Acceleration
Rearranging equation 2,
Substituting into the displacement equation,
Rearranging,
Eq. 3

10. Constant Acceleration
To find the displacement (Dx), determine the area under the v-t graph. The area is a trapezoid. The trapezoid’s area is ½(b1+b2)h, where b1 is vi and b2 is vf, and h is t.
Base 1
Height
Base 2

11. Constant Acceleration
Using the equation for the area of a trapezoid,
another equation for displacement results.
Eq. 4

12. Constant Acceleration
An equation can be obtained by squaring both sides of Equation 2.
Factoring a 2a out of the last two terms,

13. Constant Acceleration
Substituting Equation 3 for the expression in parentheses,
This results in
Eq. 5

14. 1-D Motion Equations
Eq. 1  Eq. 2  Eq. 3  Eq. 4 
Eq. 5 