Uniformly Accelerated Rectilinear Motion
Senior 2 · Physics
Uniformly Accelerated Rectilinear Motion
Lesson 3 of 4
22 min
Let's begin
Drop a stone from a bridge over the Nyabarongo and it gets faster and faster until it splashes into the water. Throw a ball straight up at break time and it slows down, stops for an instant, then falls back into your hands. Both are free fall, and the same four equations describe them.
An object is in free fall when the only force acting on it is gravity, so air resistance is small enough to ignore. Near the Earth's surface, all freely falling objects have the same acceleration, whatever their mass. This is the acceleration due to gravity, g:
g ≈ 9.8 m/s², directed downwards
Many exam questions round this to 10 m/s². Always use the value the question gives. In this lesson we use g = 9.8 m/s².
A feather falls slowly only because air resistance is large compared with its tiny weight. In a vacuum, a feather and a hammer fall side by side.
Use the same equations with a = g. Choose one direction as positive and keep to it.
A stone is dropped from rest from a bridge and hits the river 2.0 s later. Find (a) its speed as it hits the water and (b) the height of the bridge above the water.
Ishimwe throws a ball straight up at 14.7 m/s. Take upwards as positive, so a = −9.8 m/s².
At the highest point, the velocity is zero, but the acceleration is still 9.8 m/s² downwards. If the acceleration were zero there, the ball would stay in the air.
Tip
At the top: v = 0, but a = g
For a ball thrown up at speed u: time to the top = u ÷ g, and maximum height = u² ÷ 2g. Check which value of g the question gives before you start.
A stone takes 1.5 s to fall to the water at the bottom of a village well. How deep is the well to the water? (g = 9.8 m/s²)
Solution: s = ½ × 9.8 × 1.5² = 11 m (11.0 m).
A mango falls from a branch 45 m above the ground. Taking g = 10 m/s², find the time it takes to fall and its speed on landing.
Solution: 45 = ½ × 10 × t², so t² = 9.0 and t = 3.0 s. v = 10 × 3.0 = 30 m/s.
A coin is dropped from a table 1.25 m high. Using g = 10 m/s², how long does it take to reach the floor?
Solution: 1.25 = ½ × 10 × t², so t² = 0.25 and t = 0.50 s.
A ball is thrown straight up at 9.8 m/s. Using g = 9.8 m/s², find the time to the top and the maximum height.
Solution: t = 9.8 ÷ 9.8 = 1.0 s. h = 9.8² ÷ (2 × 9.8) = 4.9 m.
Summary
Sources and credits
The Rwandan examples, worked examples and practice items are written by Ganzaa.
