No-Time Acceleration Equation

v^{2} = v_{0}^{2} + 2 a s

Eliminates time from constant-acceleration kinematics. Use when two speeds and a distance are known. Displacement s is signed with the chosen axis; a is positive in that same direction.

$v$ (m/s): Final velocity
$v_{0}$ (m/s): Initial velocity
$a$ (m/s²): Constant acceleration
$s$ (m): Displacement

Step-by-step solved example

A vehicle slows from 25 m/s to 10 m/s over 75 m. Find the acceleration.

v² = v0² + 2as — the same split as braking distance.

1. Write the no-time equation
v² = v0² + 2as → 10² = 25² + 2a(75).
$100 = 625 + 150 a$
2. Solve for a
150a = 100 − 625 = −525 → a = −3.50 m/s².
$a = - 3.50 m/s^{2}$

Answer: −3.50 m/s²

10 practice questions

0/10 correct

1.From rest, a = 4 m/s² through 18 m. v =

2.A car at 20 m/s brakes at −5 m/s². Stopping distance =

3.v0 = 10 m/s, v = 30 m/s, s = 100 m. a =

4.If a = 0, the no-time equation reduces to

5.From rest, doubling the distance at the same a multiplies v by

6.v = 16 m/s from rest at a = 2 m/s². s =

7.Units of the product 2as are

8.A train slows from 30 m/s to 10 m/s in 200 m. |a| =

9.v0 = 5 m/s, a = 3 m/s², s = 8 m. v ≈

10.The equation v² = v0² + 2as assumes