When the acceleration is constant, four short equations connect displacement, the two velocities, acceleration and time. List what you know, pick the equation missing the quantity you do not need, and solve. The same kit measures g from a simple drop.
For uniformly accelerated motion in a straight line, v = u + at, s = ½(u + v)t, s = ut + ½at² and v² = u² + 2as. They follow directly from the definitions of velocity and acceleration and from the area under a velocity-time graph. Free fall is the special case with a = g.
Drop a ball from rest. The strobe marks equal time intervals, and the growing gaps are the fingerprint of uniform acceleration. The velocity-time line is straight with gradient g, exactly v = gt. Switch worlds and the value of g changes while the form of the equations stays put.
Each equation leaves one of the five quantities out. Pick the one that omits the quantity you neither know nor want.
| Equation | Leaves out | Use when you have |
|---|---|---|
| v = u + at | s | u, a, t and want v |
| s = ½(u + v)t | a | u, v, t and want s |
| s = ut + ½at² | v | u, a, t and want s |
| v² = u² + 2as | t | u, a, s and want v |
A quick derivation: from v = u + at and the area under the velocity-time graph s = ½(u + v)t, substituting one into the other removes t to give v² = u² + 2as and removes v to give s = ut + ½at².
Four quick checks on the equations and on free fall. Each correct answer earns XP and lights this skill on your star map.
The equations of uniformly accelerated motion are valid only when:
A car starts at 6.0 m s⁻¹ and accelerates at 2.0 m s⁻² for 4.0 s. Its final velocity, using v = u + at, is:
In the free-fall simulator, the velocity-time line for the falling ball is straight and passes through the origin. Its gradient is:
An object is dropped from rest. Using s = ½gt² with g = 9.8 m s⁻², how far does it fall in the first 2.0 s?
Two slips cost marks here. First, the suvat equations apply only while the acceleration is constant; a journey with separate phases must be split and each phase handled with its own values. Second, mind the signs in free fall: choose one direction as positive and stick to it. A ball thrown upward still has acceleration g acting downward throughout, including at the highest point where the velocity is momentarily zero but the acceleration is not.
Unlocks once the four checks above are done. Worth more XP, written to AS Paper 1 and 2 standard.
A car decelerates uniformly from 30 m s⁻¹ to rest over a distance of 75 m. Using v² = u² + 2as, the magnitude of the acceleration is:
A stone is dropped from rest down a deep well and takes 3.0 s to reach the water. Taking g = 9.8 m s⁻² and ignoring air resistance, the depth of the well is about:
In a g experiment a student plots fall distance s against time squared t² and obtains a straight line through the origin of gradient 4.9 m s⁻². The value of g is:
A ball is thrown vertically upward at 14 m s⁻¹. Taking g = 9.8 m s⁻² and upward as positive, its acceleration at the highest point of its flight is:
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