Kinematic Equations
Solve constant-acceleration problems using kinematic equations. Choose equations based on known and unknown quantities and interpret answers with units.
A lesson of Physics, in Measurement, Motion, and Kinematics, about 10 minutes.
By the end of this lesson, you will be able to apply the four kinematic equations to solve constant-acceleration motion problems, including free fall.
The Four Kinematic Equations
Valid for constant acceleration motion in one dimension:
$$v = v_0 + at \tag{1}$$
$$x = x_0 + v_0 t + \frac{1}{2}at^2 \tag{2}$$
$$v^2 = v_0^2 + 2a(x - x_0) \tag{3}$$
$$x = x_0 + \frac{v_0 + v}{2}t \tag{4}$$
Variables: $x$ = position, $v$ = final velocity, $v_0$ = initial velocity, $a$ = acceleration, $t$ = time.
Strategy: identify the 3 known quantities; choose the equation with only 1 unknown.
Free Fall
Free fall: motion under gravity alone (no air resistance).
$$a = -g = -9.8 \text{ m/s}^2 \quad \text{(downward)}$$
Sign convention: upward is positive. Objects thrown upward decelerate at 9.8 m/s² and reach $v = 0$ at the peak.
Symmetry of free fall: time up = time down (if thrown from and landing at same height). Speed at landing = speed at launch.
Example 1 — Free Fall
A ball is dropped from rest from a 45 m cliff. How long to hit the ground? What is the speed at impact?
$x_0 = 45$ m, $x = 0$ m, $v_0 = 0$, $a = -9.8$ m/s²
Using eq. (2): $0 = 45 + 0 - \frac{1}{2}(9.8)t^2$
$t^2 = 90/9.8 = 9.18 \implies t = 3.03$ s
Using eq. (1): $v = 0 + (-9.8)(3.03) = -29.7$ m/s (downward)
Speed = 29.7 m/s.
Example 2 — Thrown Upward
A ball is thrown upward at 20 m/s from the ground. Find: (a) maximum height, (b) time to return.
(a) At peak, $v = 0$; $v_0 = +20$ m/s, $a = -9.8$ m/s²
Eq. (3): $0 = 400 + 2(-9.8)h \implies h = 400/19.6 = 20.4$ m
(b) Back at ground: $x = x_0 = 0$, $v_0 = 20$ m/s
Eq. (2): $0 = 20t - 4.9t^2 \implies t(20 - 4.9t) = 0 \implies t = 0$ (launch) or $t = 20/4.9 = 4.08$ s
Example 3 — Car Braking
A car traveling at 30 m/s brakes with deceleration 8 m/s². How far does it travel before stopping?
$v_0 = 30$ m/s, $v = 0$, $a = -8$ m/s²
Eq. (3): $0 = 900 + 2(-8)\Delta x \implies \Delta x = 56.25$ m
Sign Convention Reminder
Always establish a positive direction at the start of the problem. Stick to it. If up is positive, then:
- $g = -9.8$ m/s²
- Upward velocity is positive
- Downward velocity is negative