Uniformly Accelerated Rectilinear Motion
Senior 2 · Physics
Uniformly Accelerated Rectilinear Motion
Lesson 2 of 4
25 min
Let's begin
If you know how fast a minibus is going and how hard its brakes can slow it, you can work out whether it will stop before a zebra crossing. Four short equations let you predict any straight-line motion with constant acceleration. Police accident investigators, driving instructors and engineers use exactly these.
u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement. Each equation links four of the five quantities and leaves one out.
| Equation | Quantity it leaves out |
|---|---|
| v = u + at | s |
| s = ½(u + v)t | a |
| s = ut + ½at² | v |
| v² = u² + 2as | t |
The first comes straight from the definition a = (v − u) ÷ t. The second says that, when acceleration is uniform, the average velocity is halfway between u and v. The last two are obtained by combining the first two.
A moto starts from rest and accelerates uniformly at 2.5 m/s² for 6.0 s. Find (a) its final velocity and (b) the distance it covers.
Known: u = 0, a = 2.5 m/s², t = 6.0 s.
A minibus travelling at 72 km/h brakes with a deceleration of 5.0 m/s². How far does it travel before stopping, and how long does it take?
First convert: u = 72 ÷ 3.6 = 20 m/s. Then v = 0 and a = −5.0 m/s². The time is not given, so use v² = u² + 2as:
A bicycle rolls down a straight stretch of a Kigali hill. It starts at 2.0 m/s and accelerates at 0.80 m/s² over 60 m. How fast is it going at the bottom?
v² = u² + 2as = 2.0² + 2 × 0.80 × 60 = 4 + 96 = 100, so v = 10 m/s.
Tip
List u, v, a, s, t, tick the three you know, and circle the one you want
Then choose the equation that contains those four and nothing else. Put a minus sign on any quantity that points opposite to your positive direction. For example, a is negative when braking.
A car moving at 5.0 m/s accelerates at 1.5 m/s² for 4.0 s. Find its final velocity and the distance covered.
Solution: v = 5.0 + 1.5 × 4.0 = 11 m/s. s = 5.0 × 4.0 + ½ × 1.5 × 4.0² = 20 + 12 = 32 m.
A moto starts from rest and reaches 24 m/s after travelling 48 m. Find its acceleration and the time taken.
Solution: v² = u² + 2as gives 576 = 0 + 2 × a × 48, so a = 6.0 m/s². Then t = (v − u) ÷ a = 24 ÷ 6.0 = 4.0 s.
A bus slows from 18 m/s to 6.0 m/s in 4.0 s. Find the deceleration and the distance covered.
Solution: a = (6.0 − 18) ÷ 4.0 = −3.0 m/s². s = ½(u + v)t = ½ × (18 + 6.0) × 4.0 = 48 m.
A minibus at 25 m/s must stop within 50 m. What deceleration is needed?
Solution: 0 = 25² + 2a × 50, so a = −625 ÷ 100 = −6.25 m/s², a deceleration of about 6.3 m/s².
Summary
Sources and credits
The Rwandan examples, worked examples and practice items are written by Ganzaa.
