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KinematicsPhysicsHigh schoolabout 0:49

A car moving at 18 m/s brakes at 4 m/s² until it stops. Find the stopping distance and time: animated lesson

A narrated, step-by-step kinematics video lesson with captions. Final answer: s = 40.5 m, t = 4.5 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 (0.3 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
A car moving at 18 m/s brakes at 4 m/s² until it stops. Find the stopping distance and time.
2. Step by step
1. $u = 18\ \text{m/s},\quad v = 0\ \text{m/s},\quad a = -4\ \text{m/s}^2$
  List the known quantities
2. $t = \frac{v - u}{a} = 4.5$
  Time
3. $v^2 = u^2 + 2as \Rightarrow s = \frac{v^2 - u^2}{2a} = 40.5$
  Displacement
3. The answer
$s = 40.5\ \text{m},\ t = 4.5\ \text{s}$
s = 40.5 m, t = 4.5 s
4. Check
- Check v² = u² + 2as: 0 = 0.

Transcript

The problem. Here is the problem: A car moving at 18 m/s brakes at 4 m/s squared until it stops. Find the stopping distance and time. Find displacement s and time t for motion in a straight line with constant acceleration.

Step by step. List the known quantities. Unknown: displacement s and time t. Time. Displacement.

The answer. So the answer is s equals 40.5 m, t equals 4.5 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: 0 equals 0.

About kinematics

Motion with constant acceleration: distance, velocity and time.

A car moving at 18 m/s brakes at 4 m/s² until it stops.…