Maths › Mechanics › Kinematics with constant acceleration
Kinematics with constant acceleration
When acceleration holds steady, five quantities lock together and any three determine the rest. The suvat equations are the bookkeeping, the velocity-time graph is the reason they work, and gravity is the standard example.
Builds on Modelling, quantities and units.
IN THIS TOPIC
- Read displacement and acceleration from velocity-time graphs.
- Select and use the suvat equations for constant acceleration.
- Model vertical motion under gravity, using the symmetry of up-and-down flight.
WHAT YOU PROBABLY THINK
At the top of its flight a thrown ball has zero acceleration.
The graph behind the formulas
Plot velocity against time and constant acceleration is a straight line: gradient is the acceleration, area underneath is the displacement. Every suvat equation is this trapezium described in symbols:
Each equation omits exactly one of the five letters. The working method is clerical and reliable: list s, u, v, a, t; fill in the three you know; pick the equation missing the one you do not care about.
WORKED EXAMPLE
A braking train
A train slows from 24 m s⁻¹ to 12 m s⁻¹ over 300 m. Find the deceleration.
Known: u = 24, v = 12, s = 300; wanted: a; time not involved, so use v² = u² + 2as.
144 = 576 + 600a, so a = −432/600 = −0.72 m s⁻²: a deceleration of 0.72.
The sign did the physics: negative because the train slows while moving in the positive direction.
Gravity: the standard constant
Free flight near the ground is constant acceleration g = 9.8 m s⁻² downwards, air resistance modelled away. Taking up as positive, a = −9.8 for the whole flight, on the way up, at the top, and on the way down. At the peak the velocity is zero; the acceleration never is.
A ball thrown up at 14.7 m s⁻¹ reaches v = 0 when t = 14.7/9.8 = 1.5 s, at height 14.7 × 1.5 − 4.9 × 1.5² = 11.0 m, and lands with speed 14.7 again: the flight is symmetric, and the symmetry is a free answer-checker.
THE EXAM BIT
- Write the suvat list and fill it in before choosing an equation; the selection is then automatic.
- suvat applies only while acceleration is constant: split the motion at the moment the acceleration changes.
- In vertical motion, declare the positive direction and keep g's sign consistent with it.
- Graph questions pay for areas and gradients, not for formulas: say which you are using.
CHECK YOURSELF
A stone is dropped from rest down a well and hits the water after 2 s. Taking g = 9.8 m s⁻², find the depth and the speed at impact.
Show a hint
u = 0 makes two suvat equations very short.
Show the answer
Depth: s = ½ × 9.8 × 2² = 19.6 m.
Impact speed: v = 9.8 × 2 = 19.6 m s⁻¹.
List s, u, v, a, t; pick the equation missing the letter you do not need.
Gravity is a constant −9.8 with up positive, all flight long, peak included.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Read displacement and acceleration from velocity-time graphs.
- Select and use the suvat equations for constant acceleration.
- Model vertical motion under gravity, using the symmetry of up-and-down flight.
No animated video for this topic yet; these notes stand alone.