Uniformly Accelerated Rectilinear Motion
Senior 2 · Physics
Uniformly Accelerated Rectilinear Motion
Lesson 1 of 4
18 min
Let's begin
When the light turns green at a junction in Kigali, a moto taxi pulls away and gets faster every second. When a minibus driver sees a zebra crossing, the bus slows down steadily. Both are examples of accelerated motion. Road safety engineers use this topic to set speed limits and stopping distances, and it is the foundation for everything else in mechanics.
A moto starts from rest (u = 0) on a straight road and reaches 12 m/s after 4.0 s. What is its acceleration?
a = (v − u) ÷ t = (12 − 0) ÷ 4.0 = 3.0 m/s². Its velocity rises by 3.0 m/s every second: 3, 6, 9 and then 12 m/s.
Motion is uniformly accelerated when the velocity changes by equal amounts in equal time intervals, so the acceleration is constant. Uniformly accelerated rectilinear motion is therefore motion in a straight line with constant acceleration. Its characteristics are:
If the velocity does not change at all, the acceleration is zero. This is uniform motion (constant velocity), a special case with a = 0.
A minibus moving at 20 m/s brakes steadily and stops in 5.0 s.
a = (0 − 20) ÷ 5.0 = −4.0 m/s². The minus sign shows that the acceleration is opposite to the motion, so the bus is slowing down. A negative acceleration like this is often called a deceleration (or retardation) of 4.0 m/s².
Speeds on road signs are in km/h. To convert, divide by 3.6, because 1 km/h = 1000 m ÷ 3600 s. For example, 72 km/h = 72 ÷ 3.6 = 20 m/s, and 54 km/h = 15 m/s.
Open the Forces and Motion: Basics simulation and choose the Acceleration screen. Apply a steady push to the cart with friction turned off, and watch the speedometer. The speed rises by the same amount every second, which is uniform acceleration.
Tip
a = (v − u) ÷ t: "final minus first, over time"
Always convert km/h to m/s (÷ 3.6) before you use this formula. A negative answer means the object is slowing down in the direction you chose as positive.
A cyclist speeds up from 2.0 m/s to 8.0 m/s in 3.0 s on a straight road. Find the acceleration.
Solution: a = (8.0 − 2.0) ÷ 3.0 = 2.0 m/s².
A lorry at 15 m/s brakes to a stop in 6.0 s. Find its acceleration.
Solution: a = (0 − 15) ÷ 6.0 = −2.5 m/s², which is a deceleration of 2.5 m/s².
Convert 90 km/h to m/s.
Solution: 90 ÷ 3.6 = 25 m/s.
A car moves along a straight road at a steady 20 m/s for one minute. What is its acceleration, and what kind of motion is this?
Solution: The velocity does not change, so a = 0 m/s². This is uniform motion (constant velocity).
Summary
Sources and credits
The Rwandan examples, worked examples and practice items are written by Ganzaa.
