Kinematics Equations Explained (How to Pick One)

Symbol	Meaning	Units
$\Delta x$	displacement (change in position)	m
$v_0$	initial velocity	m/s
$v$	final velocity	m/s
$a$	acceleration (must be constant)	$\text{m/s}^2$
$t$	time	s

Every problem gives you three of these and asks for a fourth. Watch for hidden values in the wording:

• "Starts from rest" means $v_0 = 0$.
• "Comes to a stop" means $v = 0$.
• "Dropped" means $v_0 = 0$ and $a = -9.8 \text{ m/s}^2$ (with up as positive).
• "At the highest point" means $v = 0$ for a vertical throw.

$$\text{(1)} \quad v = v_0 + at$$ $$\text{(2)} \quad \Delta x = v_0t + \tfrac{1}{2}at^2$$ $$\text{(3)} \quad v^2 = v_0^2 + 2a\Delta x$$ $$\text{(4)} \quad \Delta x = \tfrac{1}{2}(v_0 + v)t$$

How to pick one: each equation is missing exactly one variable.

Equation	Missing variable
(1) $v = v_0 + at$	$\Delta x$
(2) $\Delta x = v_0t + \frac{1}{2}at^2$	$v$
(3) $v^2 = v_0^2 + 2a\Delta x$	$t$
(4) $\Delta x = \frac{1}{2}(v_0 + v)t$	$a$

1. List the three known variables and the one you want.
2. Find the variable that is neither given nor asked for.
3. Use the equation that is missing that variable.

Equation (1), no $\Delta x$: A car going 12 m/s accelerates at 3.0 $\text{m/s}^2$ for 5.0 s. How fast is it going?

Known: $v_0 = 12$, $a = 3.0$, $t = 5.0$. Want: $v$. Missing: $\Delta x$.

$$v = 12 + (3.0)(5.0) = 27 \text{ m/s}$$

Equation (2), no $v$: A bike starts from rest and accelerates at 2.5 $\text{m/s}^2$ for 8.0 s. How far does it go?

Known: $v_0 = 0$, $a = 2.5$, $t = 8.0$. Want: $\Delta x$. Missing: $v$.

$$\Delta x = 0 + \tfrac{1}{2}(2.5)(8.0)^2 = \tfrac{1}{2}(2.5)(64) = 80. \text{ m}$$

Equation (3), no $t$: A car moving at 25.0 m/s brakes to a stop in 50.0 m. What is its acceleration?

Known: $v_0 = 25.0$, $v = 0$, $\Delta x = 50.0$. Want: $a$. Missing: $t$.

$$0 = (25.0)^2 + 2a(50.0) \quad\Rightarrow\quad 0 = 625 + 100a \quad\Rightarrow\quad a = -6.25 \text{ m/s}^2$$

The negative sign means the acceleration points opposite the motion, so the car is slowing down.

Equation (4), no $a$: A train slows from 30.0 m/s to 10.0 m/s in 20.0 s. How far does it travel?

Known: $v_0 = 30.0$, $v = 10.0$, $t = 20.0$. Want: $\Delta x$. Missing: $a$.

$$\Delta x = \tfrac{1}{2}(30.0 + 10.0)(20.0) = \tfrac{1}{2}(40.0)(20.0) = 400. \text{ m}$$

An object in free fall has constant acceleration $g = 9.8 \text{ m/s}^2$ downward. Just replace $x$ with $y$ and pick a sign convention.

Example 1: A rock is dropped from a 45.0 m cliff. How long does it fall, and how fast is it going at the bottom?

Take down as positive here, so $a = +9.8$, $v_0 = 0$, and $\Delta y = 45.0$ m.

1. Time, using equation (2): $45.0 = \frac{1}{2}(9.8)t^2$, so $t^2 = 9.18$ and $t = 3.03$ s.
2. Speed, using equation (3): $v^2 = 0 + 2(9.8)(45.0) = 882$, so $v = 29.7$ m/s.

Example 2: A ball is thrown straight up at 15.0 m/s. How high does it go?

Take up as positive, so $a = -9.8$. At the top, $v = 0$. Time is not needed, so use equation (3):

$$0 = (15.0)^2 + 2(-9.8)\Delta y \quad\Rightarrow\quad \Delta y = \frac{225}{19.6} = 11.5 \text{ m}$$

Time to the top, from equation (1): $0 = 15.0 - 9.8t$, so $t = 1.53$ s.

Signs. Pick a positive direction at the start and stick with it. Velocity, displacement, and acceleration all take signs from that choice.

• If up is positive, gravity is $-9.8 \text{ m/s}^2$ the whole time, even while the object is rising.
• A negative acceleration does not always mean "slowing down." An object speeds up when velocity and acceleration have the same sign and slows down when they have opposite signs.

Multi-part problems. The equations only work while acceleration is constant. If a car accelerates and then brakes, split the trip into two parts. The final velocity of part one becomes the initial velocity of part two.

Paste it into the VeilGPT solver to check your steps on a multi-part problem.