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KinematicsPhysicsHigh schoolabout 0:56

An object speeds up from 5 m/s to 25 m/s over 200 m. Find its acceleration and the time taken: animated lesson

A narrated, step-by-step kinematics video lesson with captions. Final answer: a = 1.5 m/s², t = 13.333333 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 speeds up from 5 m/s to 25 m/s over 200 m. Find its acceleration and the time taken.
2. Step by step
1. $s = 200\ \text{m},\quad u = 5\ \text{m/s},\quad v = 25\ \text{m/s}$
  List the known quantities
2. $s = \tfrac12 (u + v)t \Rightarrow t = \frac{2s}{u + v} = 13.333333$
  Time from the average velocity
3. $a = \frac{v - u}{t} = 1.5$
  Acceleration
3. The answer
$a = 1.5\ \text{m/s}^2,\ t = 13.333333\ \text{s}$
a = 1.5 m/s², t = 13.333333 s
4. Check
- Check v² = u² + 2as: 625 = 625.

Transcript

The problem. Here is the problem: An object speeds up from 5 m/s to 25 m/s over 200 m. Find its acceleration and the time taken. Find acceleration a and time t for motion in a straight line with constant acceleration.

Step by step. List the known quantities. Unknown: acceleration a and time t. Time from the average velocity. Acceleration.

The answer. So the answer is a equals 1.5 m/s squared, t equals 13.333333 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: 625 equals 625.

About kinematics

Motion with constant acceleration: distance, velocity and time.

An object speeds up from 5 m/s to 25 m/s over 200 m. Find…