Notes
Describing motion
- Scalars have only size (distance, speed, time). Vectors have size and direction (displacement, velocity, acceleration).
- Displacement $\Delta x = x_f - x_i$. Average velocity $\displaystyle = \frac{\Delta x}{\Delta t}$. Acceleration $\displaystyle = \frac{\Delta v}{\Delta t}$.
- An object slows down when velocity and acceleration point in opposite directions.
Constant-acceleration equations
- Free fall: $a = -g$, with $g \approx 9.8$ m/s² (often rounded to 10 m/s²). At the top of a toss, $v = 0$ but $a$ is still $-g$.
Graphs
| Graph | Slope gives | Area gives |
|---|---|---|
| Position vs time | Velocity | — |
| Velocity vs time | Acceleration | Displacement |
| Acceleration vs time | — | Change in velocity |
Two-dimensional motion and projectiles
- Treat horizontal and vertical motion separately. They share only time.
- With no air resistance, horizontal velocity stays constant; vertical motion has $a_y = -g$.
- A ball launched horizontally from height $h$ lands at the same time as one dropped from rest from $h$.
- Launch angle $\theta$: $v_x = v\cos\theta$, $v_y = v\sin\theta$.
Reference frames
Velocity depends on the observer. Relative velocity: $\vec v_{A \text{ rel } C} = \vec v_{A \text{ rel } B} + \vec v_{B \text{ rel } C}$.
Practice
$\displaystyle a = \frac{20 - 0}{4.0} = 5.0$ m/s².
Gravity keeps acting the whole time.
$\Delta x = v\,\Delta t$, which is the area.
$\displaystyle \frac{1}{2}(10)(3.0)^2 = 45$ m.
Both start with zero vertical velocity and have the same vertical acceleration.
$20\cos 30° \approx 17.3$ m/s.
$v^2 = v_0^2 + 2a\Delta x$: $25 = 625 + 120a$, so $a = -5.0$ m/s².
Velocities in the same direction add.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A cart starts from rest and accelerates uniformly at 2.0 m/s² for 5.0 s, then moves at constant velocity for 3.0 s.
(a) Find the cart's velocity at $t = 5.0$ s.
Show the answer and scoring guide
$v = 0 + (2.0)(5.0) = 10$ m/s.
Check off the points you earned:
(b) Find the total displacement over the 8.0 s.
Show the answer and scoring guide
First 5 s: $\displaystyle \frac{1}{2}(2.0)(5.0)^2 = 25$ m. Next 3 s: $10 \times 3.0 = 30$ m. Total: 55 m.
Check off the points you earned:
(c) Describe the velocity-time graph for the 8.0 s.
Show the answer and scoring guide
A straight line from (0, 0) rising to (5.0 s, 10 m/s) with slope 2.0 m/s², then a horizontal line at 10 m/s from 5.0 s to 8.0 s.
Check off the points you earned: