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KinematicsPhysicsHigh schoolabout 0:57

An object starts at 5 m/s and accelerates at 2 m/s² for 4 s. How far does it go and how fast is it moving?: animated lesson

A narrated, step-by-step kinematics video lesson with captions. Final answer: s = 36 m, v = 13 m/s.

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Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.4 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
An object starts at 5 m/s and accelerates at 2 m/s² for 4 s. How far does it go and how fast is it moving?
2. Step by step
1. $u = 5\ \text{m/s},\quad a = 2\ \text{m/s}^2,\quad t = 4\ \text{s}$
  List the known quantities
2. $v = u + at = 5 + (2)(4) = 13$
  Final velocity
3. $s = ut + \tfrac12at^2 = 36$
  Displacement
3. The answer
$s = 36\ \text{m},\ v = 13\ \text{m/s}$
s = 36 m, v = 13 m/s
4. Check
- Check v² = u² + 2as: 169 = 169.

Transcript

The problem. Here is the problem: An object starts at 5 m/s and accelerates at 2 m/s squared for 4 s. How far does it go and how fast is it moving? Find displacement s and final velocity v for motion in a straight line with constant acceleration.

Step by step. List the known quantities. Unknown: displacement s and final velocity v. Final velocity. Displacement.

The answer. So the answer is s equals 36 m, v equals 13 m/s. With constant acceleration, any three of displacement, initial velocity, final velocity, acceleration and time fix the other two. Each pair of unknowns is solved with the equation that does not contain the other unknown.

Check. Check v squared equals u squared plus 2as: 169 equals 169.

About kinematics

Motion with constant acceleration: distance, velocity and time.

An object starts at 5 m/s and accelerates at 2 m/s² for 4…