Motion in one dimension and free fall

Kinematics describes motion without looking at its cause: where the object is, how fast it is going, how its velocity changes. This week we are in one dimension: the motion is along a line, and direction is shown only by a $+$ or $-$ sign. The midterm typically has 4 questions: a derivative from $x(t)$, braking, free fall, a $v$–$t$ graph.

Golden rules:

1. First choose an axis and a positive direction. In free fall usually up is $+$, so $a = -g$.
2. Of the four constant-acceleration equations, choose by the missing quantity. If time is not asked, use $v^2 = v_0^2 + 2a\Delta x$.
3. Zero velocity does not mean zero acceleration. At the top, $v = 0$ and $a = -g$.
4. The area under the $v$–$t$ graph is the displacement.
5. Convert the units: km/h does not go into a formula.

Position, displacement, distance

Position $x$ is the object’s place relative to the origin you choose. Displacement $\Delta x = x_f - x_i$ is a signed quantity. Distance is the total length traveled, always positive.

Example: a runner who goes 120 m east, then 40 m west. $\Delta x = +80$ m, distance = 160 m.

Average and instantaneous velocity

$\bar v = \frac{\Delta x}{\Delta t} \qquad \text{average speed} = \frac{\text{total distance}}{\Delta t}$

Instantaneous velocity is the average velocity in the limit $\Delta t \to 0$, that is, the derivative of position:

$v = \frac{dx}{dt}$

On a graph: the slope of the tangent to the $x$–$t$ curve at a point is the velocity at that moment. Speed is the magnitude of the velocity, $|v|$.

Acceleration

$\bar a = \frac{\Delta v}{\Delta t} \qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \qquad [a] = \text{m/s}^2$

The sign of acceleration is not the answer to “is it speeding up?”. If $a$ and $v$ have the same sign, the object speeds up; if opposite signs, it slows down. An object moving in the negative direction with negative acceleration is speeding up.

Example 1. $x(t) = 2t^2 - 8t + 3$ (m, s). When does it stop? What are the displacement and distance between 0 and 4 s?

v = dx/dt = 4t - 8          v = 0  →  t = 2 s (turning point)
a = dv/dt = 4 m/s²          (constant)
x(0) = 3 m,  x(2) = -5 m,  x(4) = 3 m
Δx (0→4) = 3 - 3 = 0
distance = |(-5) - 3| + |3 - (-5)| = 8 + 8 = 16 m

To find the distance in motion that changes direction, split at the turning points. $x(4) - x(0)$ gives you only the displacement.

Constant-acceleration motion

When $a$ is constant there are four equations. Each is missing one of the five quantities ($\Delta x, v_0, v, a, t$):

Method: list what is known and what is asked, and choose the equation that does not contain the quantity not in the list.

Example 2. A car going 25 m/s slows down at $4.0\,\text{m/s}^2$. In how many seconds and how many meters does it stop?

Known: v0 = 25 m/s, v = 0, a = -4.0 m/s²
t:   0 = 25 + (-4.0) t        →  t = 6.25 s
Δx:  0 = 25² + 2(-4.0) Δx     →  Δx = 625 / 8.0 = 78.1 m
check: Δx = ½ (25 + 0)(6.25) = 78.1 m

PITFALL · the sign of acceleration in braking. If the direction of motion is $+$, the $a$ of a slowing object is negative. If you put $a = +4.0$, you get $v^2 = 625 + 8\Delta x$ and $\Delta x$ comes out negative. The moment you see a negative distance, check the sign.

Two-body problems

In catch-up questions, write a separate $x(t)$ for each object, with the same origin and the same clock. Then solve $x_A(t) = x_B(t)$. The moment the velocities are equal is not the catch-up moment; it is the moment the gap is largest (or smallest).

Reading graphs

In a $v$–$t$ graph, the area below the axis is negative displacement. If the distance is wanted, add the absolute values of the areas.

Free fall

Neglecting air resistance, every object near the Earth’s surface falls with the same acceleration: $g = 9.80\,\text{m/s}^2$, downward. It is independent of mass. With up chosen as $+$, the constant-acceleration equations are used with $a = -g$:

$v = v_0 - gt \qquad y = y_0 + v_0t - \tfrac12gt^2 \qquad v^2 = v_0^2 - 2g(y - y_0)$

Ready results for an object thrown up ($v_0$ upward):

$t_{\text{top}} = \frac{v_0}{g} \qquad H = \frac{v_0^2}{2g} \qquad t_{\text{return}} = \frac{2v_0}{g}$

When it returns to the launch point, the speed is $v_0$ again, directed downward. The motion is symmetric.

Example 3. A ball is thrown upward at 15.0 m/s from a roof 20.0 m high and falls to the ground below. What are the peak height, flight time and impact velocity?

Up +, roof y = 0, ground y = -20.0 m, a = -9.80 m/s²
t_top = 15.0 / 9.80 = 1.53 s
H     = 15.0² / (2 × 9.80) = 11.5 m (above the roof)
Ground:  -20.0 = 15.0 t - 4.90 t²   →   4.90 t² - 15.0 t - 20.0 = 0
t = [15.0 + √(225 + 392)] / 9.80 = 4.07 s   (the negative root is discarded)
v = 15.0 - 9.80 × 4.07 = -24.8 m/s   (downward)
check: v² = 15.0² + 2(9.80)(20.0)  →  |v| = 24.8 m/s

PITFALL · “dropped” vs “thrown down”. “Dropped” means $v_0 = 0$. “Thrown downward at 8 m/s” means $v_0 = -8$ m/s on an up-is-$+$ axis. If you mix up the sign, the root of the quadratic comes out completely different, and that wrong answer is waiting ready among the options.

Common mistakes