The kinematics equations: how to pick and use them

Q: Why is g sometimes 9.81 or 10?

The real value near Earth's surface is about 9.81\ \text{m/s}^2 and changes a little from place to place. Many classes round to 9.8, and some use 10 to make the math quick. Use the value your teacher or test gives, and keep it the same throughout a problem.

Key ideas

v = v_0 + at

\Delta x = v_0 t + \tfrac{1}{2} a t^2

v^2 = v_0^2 + 2a\,\Delta x

\Delta x = \tfrac{1}{2}(v_0 + v)\,t

$\Delta x$ is displacement, the change in position, in meters (m). It can be negative.
$v_0$ is the initial velocity and $v$ is the final velocity, in meters per second (m/s).
$a$ is the acceleration, in meters per second squared (m/s²). It must be constant for these equations to work.
$t$ is the time, in seconds (s).

Pick the equation by the quantity you don't have
Equation	Leaves out	Use it when
$v = v_0 + at$	$\Delta x$	No distance is given or asked
$\Delta x = v_0 t + \tfrac{1}{2}at^2$	$v$	No final velocity is given or asked
$v^2 = v_0^2 + 2a\Delta x$	$t$	No time is given or asked
$\Delta x = \tfrac{1}{2}(v_0 + v)t$	$a$	No acceleration is given or asked

Free fall is the same math in the vertical direction. If you take up as positive, $a = -g = -9.8\ \text{m/s}^2$ the whole time an object is in the air, on the way up, at the top and on the way down.

Worked examples

Example 1: speeding up from rest

Problem A car starts from rest and speeds up at 3.0 m/s² for 5.0 s. How fast is it going, and how far does it travel?

Knowns: $v_0 = 0$ , $a = 3.0\ \text{m/s}^2$ , $t = 5.0\ \text{s}$ .
Final speed, using the equation without $\Delta x$ .
$v = 0 + (3.0\ \text{m/s}^2)(5.0\ \text{s}) = 15\ \text{m/s}$
Distance, using the equation without $v$ .
$\Delta x = 0 + \tfrac{1}{2}(3.0\ \text{m/s}^2)(5.0\ \text{s})^2 = 37.5\ \text{m}$
The data have 2 significant figures, so round to 38 m.

Answer 15 m/s after traveling 38 m

Example 2: braking distance

Problem A car moving at 24 m/s brakes with an acceleration of $-5.0\ \text{m/s}^2$ . How far does it go before it stops, and how long does that take?

Knowns: $v_0 = 24\ \text{m/s}$ , $v = 0$ , $a = -5.0\ \text{m/s}^2$ . No time given, so use $v^2 = v_0^2 + 2a\Delta x$ .
$0 = (24\ \text{m/s})^2 + 2(-5.0\ \text{m/s}^2)\,\Delta x$
Solve for $\Delta x$ .
$\Delta x = \frac{576\ \text{m}^2/\text{s}^2}{10\ \text{m/s}^2} = 57.6\ \text{m}$
Time, from $v = v_0 + at$ .
$t = \frac{0 - 24\ \text{m/s}}{-5.0\ \text{m/s}^2} = 4.8\ \text{s}$

Answer 58 m, in 4.8 s

Example 3: dropping a ball

Problem A ball is dropped from a window 20.0 m above the ground. How long does it fall, and how fast is it going when it lands? Ignore air resistance.

Take down as positive this time so every value is positive: $v_0 = 0$ , $a = 9.8\ \text{m/s}^2$ , $\Delta x = 20.0\ \text{m}$ .
Time, from $\Delta x = \tfrac{1}{2}at^2$ .
$t = \sqrt{\frac{2\Delta x}{a}} = \sqrt{\frac{2(20.0\ \text{m})}{9.8\ \text{m/s}^2}} = 2.020\ \text{s}$
Landing speed.
$v = at = (9.8\ \text{m/s}^2)(2.020\ \text{s}) = 19.80\ \text{m/s}$

Answer 2.02 s, landing at 19.8 m/s

Example 4: throwing straight up

Problem A ball is thrown straight up at 15.0 m/s. How high does it go, and how long does it take to reach the top?

Up is positive: $v_0 = 15.0\ \text{m/s}$ , $a = -9.8\ \text{m/s}^2$ , and $v = 0$ at the top.
Height, from the equation without time.
$\Delta x = \frac{v^2 - v_0^2}{2a} = \frac{0 - (15.0\ \text{m/s})^2}{2(-9.8\ \text{m/s}^2)} = 11.48\ \text{m}$
Time to the top.
$t = \frac{v - v_0}{a} = \frac{0 - 15.0\ \text{m/s}}{-9.8\ \text{m/s}^2} = 1.531\ \text{s}$

Answer 11.5 m high, reached after 1.53 s

Common mistakes and how to fix them

Mixing signs. If up is positive, gravity is $-9.8\ \text{m/s}^2$ , even when the ball moves up. Fix: write your positive direction at the top of the page.
Using the equations when acceleration changes. They only work for constant acceleration. Fix: split the motion into parts where $a$ is constant.
Thinking acceleration is zero at the top. Velocity is zero at the top, but gravity still acts, so $a = -9.8\ \text{m/s}^2$ .
Forgetting to square the time. In $\tfrac{1}{2}at^2$ , square $t$ before you multiply.
Mixing units. Fix: convert km/h to m/s and minutes to seconds before you start.

Practice problems

A bike moving at 2.0 m/s speeds up at 1.5 m/s² for 4.0 s. What is its final speed?
1. 6.0 m/s
2. 8.0 m/s
3. 14 m/s
4. 3.5 m/s
Show answer
Answer: 8.0 m/s
$v = v_0 + at = 2.0 + (1.5)(4.0) = 8.0$ m/s. 6.0 m/s forgets the starting speed.
A plane starts from rest, accelerates at 3.2 m/s², and needs 64 m/s to take off. What is the shortest runway it can use?
Show answer
Answer: 640 m
No time is given, so use $v^2 = v_0^2 + 2a\Delta x$ : $\Delta x = \frac{64^2 - 0}{2(3.2)} = 640$ m.
Which equation should you use if a problem gives no time and asks for no time?
1. $v = v_0 + at$
2. $\Delta x = v_0 t + \tfrac{1}{2}at^2$
3. $v^2 = v_0^2 + 2a\Delta x$
4. $\Delta x = \tfrac{1}{2}(v_0 + v)t$
Show answer
Answer: $v^2 = v_0^2 + 2a\Delta x$
It is the only one of the four with no $t$ in it.
A stone is dropped from a bridge and hits the water 3.0 s later. How high is the bridge? Ignore air resistance.
Show answer
Answer: 44 m
$\Delta x = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.8)(3.0)^2 = 44.1$ m, which is 44 m to 2 significant figures.
A car slows from 30. m/s to 10. m/s in 4.0 s. What is its acceleration?
1. -5.0 m/s²
2. 5.0 m/s²
3. -20 m/s²
4. -2.5 m/s²
Show answer
Answer: -5.0 m/s²
$a = \frac{v - v_0}{t} = \frac{10 - 30}{4.0} = -5.0$ m/s². It is negative because the car is slowing down while moving in the positive direction.

Frequently asked questions

What is the difference between speed and velocity?

Speed is how fast something moves. Velocity is speed plus a direction, so it can be negative. A car going around a curve at a steady 20 m/s has constant speed but changing velocity. The kinematics equations use velocity, so the signs matter.

Can I use these equations for objects thrown at an angle?

Yes, one direction at a time. Split the motion into horizontal and vertical parts and use the equations for each. Horizontally $a = 0$ , and vertically $a = -9.8\ \text{m/s}^2$ . See projectile motion.

Why is g sometimes 9.81 or 10?

The real value near Earth's surface is about $9.81\ \text{m/s}^2$ and changes a little from place to place. Many classes round to 9.8, and some use 10 to make the math quick. Use the value your teacher or test gives, and keep it the same throughout a problem.

Sources

OpenStax College Physics 2e, 2.5 Motion Equations for Constant Acceleration in One Dimension, accessed October 1, 2026
OpenStax College Physics 2e, 2.7 Falling Objects, accessed October 1, 2026

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

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