Equations of Motion and Terminal Velocity

Drop a coin and a feather together and the coin hits the floor first — but only because of the air. Take the air away (as in a famous vacuum-tube demonstration, and on the airless Moon) and the two fall side by side, hitting the ground at the same instant. This is because, near the Earth's surface, every falling object accelerates at the same rate under gravity. To handle these falling and accelerating objects we need one more equation linking velocity, acceleration and distance — and we need to understand why a skydiver, after speeding up at first, eventually falls at a steady terminal velocity. This lesson, part of Topic P2 (Forces) of OCR Gateway Science A, introduces the equation $v^2 = u^2 + 2as$v2=u2+2as, explains free fall under gravity, and accounts for terminal velocity using a velocity–time graph.

By the end of this lesson you should be able to use the equation $v^2 = u^2 + 2as$v2=u2+2as, describe free fall with a gravitational field strength of $g = 9.8\text{ N/kg}$g=9.8 N/kg, and explain terminal velocity in terms of balanced forces using a velocity–time graph.

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The Equation $v^2 = u^2 + 2as$v2=u2+2as

For an object moving with a constant (uniform) acceleration, the final velocity, starting velocity, acceleration and distance are linked by:

$v^2 = u^2 + 2as$v2=u2+2as

where $u$u is the starting (initial) velocity (in $\text{m/s}$m/s), $v$v is the final velocity (in $\text{m/s}$m/s), $a$a is the acceleration (in $\text{m/s}^2$m/s2) and $s$s is the distance travelled (in $\text{m}$m).

Higher tier: Using this equation in calculations is Higher tier content. The good news is that OCR provides this equation on the equation sheet in the exam, so you do not have to memorise it — but you must be able to select it, rearrange it and use it correctly.

This equation is especially useful because it has no time in it — so you can use it whenever you know the velocities, the acceleration and the distance but not the time. Rearranged forms include:

$v^2 = u^2 + 2as \qquad a = \dfrac{v^2 - u^2}{2s} \qquad s = \dfrac{v^2 - u^2}{2a}$v2=u2+2asa=2sv2−u2​s=2av2−u2​

Worked Example 1 — Finding a final velocity (Higher)

A car accelerates from rest ($u = 0 \text{ m/s}$u=0 m/s) at $3 \text{ m/s}^2$3 m/s2 over a distance of $24 \text{ m}$24 m. Find its final velocity.

Step 1 — write the equation: $v^2 = u^2 + 2as$v2=u2+2as.

Step 2 — substitute: $v^2 = 0^2 + 2 \times 3 \times 24$v2=02+2×3×24.

Step 3 — work out the right-hand side: $v^2 = 0 + 144 = 144$v2=0+144=144.

Step 4 — take the square root: $v = \sqrt{144} = 12 \text{ m/s}$v=144​=12 m/s.

Answer: the final velocity is $12 \text{ m/s}$12 m/s.

Worked Example 2 — Finding an acceleration (Higher)

A train slows from $30 \text{ m/s}$30 m/s to $10 \text{ m/s}$10 m/s over a distance of $400 \text{ m}$400 m. Calculate its acceleration.

Step 1 — rearrange for $a$a: $a = \dfrac{v^2 - u^2}{2s}$a=2sv2−u2​.

Step 2 — substitute: $a = \dfrac{10^2 - 30^2}{2 \times 400} = \dfrac{100 - 900}{800}$a=2×400102−302​=800100−900​.

Step 3 — calculate: $a = \dfrac{-800}{800} = -1 \text{ m/s}^2$a=800−800​=−1 m/s2.

Answer: the acceleration is $-1 \text{ m/s}^2$−1 m/s2. The minus sign confirms the train is decelerating.

Exam Tip: (Higher) $v^2 = u^2 + 2as$v2=u2+2as is on the equation sheet — find it there rather than guessing. Use it when time is not given. Remember to square $u$u and $v$v, and to take the square root at the end when finding $v$v.

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Free Fall and Gravitational Field Strength

Near the Earth's surface, gravity pulls every object downward, giving it a downward acceleration. The strength of this pull is described by the gravitational field strength, $g$g, which on Earth is about:

$g = 9.8\text{ N/kg}$g=9.8 N/kg

The gravitational field strength tells you the force of gravity on each kilogram of mass — that is what "$\text{N/kg}$N/kg" means, newtons per kilogram. The same number, $9.8$9.8, is also the acceleration of free fall in $\text{m/s}^2$m/s2: in the absence of air resistance, every object falls with the same acceleration of about $9.8 \text{ m/s}^2$9.8 m/s2, regardless of its mass. (You may see $g$g rounded to $10 \text{ N/kg}$10 N/kg in some questions to make the arithmetic easier.)

This is why, in a vacuum, a coin and a feather fall together: with no air resistance, the only force is gravity, and gravity gives them the same acceleration. On the Moon, where there is no atmosphere and $g$g is only about $1.6 \text{ N/kg}$1.6 N/kg, objects still fall together but much more slowly — famously demonstrated by an astronaut dropping a hammer and a feather.

Exam Tip: On Earth $g = 9.8\text{ N/kg}$g=9.8 N/kg — this is both the force per kilogram and the acceleration of free fall. Without air resistance, all objects fall at the same rate whatever their mass, because gravity gives them the same acceleration.

Worked Example 3 — A falling object (Higher)

A stone is dropped from rest down a well and falls $20 \text{ m}$20 m before hitting the water. Taking $g = 9.8 \text{ m/s}^2$g=9.8 m/s2 and ignoring air resistance, find the speed at which it hits the water.

Step 1 — choose the equation (we know $u$u, $a$a and $s$s but not time): $v^2 = u^2 + 2as$v2=u2+2as.

Step 2 — substitute ($u = 0$u=0, $a = 9.8$a=9.8, $s = 20$s=20): $v^2 = 0^2 + 2 \times 9.8 \times 20$v2=02+2×9.8×20.

Step 3 — calculate: $v^2 = 392$v2=392.

Step 4 — square root: $v = \sqrt{392} = 19.8 \text{ m/s}$v=392​=19.8 m/s (to 3 significant figures).

Answer: the stone hits the water at about $19.8 \text{ m/s}$19.8 m/s.

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Terminal Velocity

When an object falls through air (or any fluid), two forces act on it: its weight pulling it down, and air resistance (drag) pushing up. Air resistance is special because it increases with speed — the faster the object moves, the bigger the drag. This leads to a steady final speed called the terminal velocity.

Follow a skydiver after they jump:

1. At the start, the skydiver is moving slowly, so the air resistance is small. The weight is much greater than the air resistance, so there is a large resultant force downward, and the skydiver accelerates rapidly.
2. As the speed increases, the air resistance increases. The resultant force (weight − air resistance) gets smaller, so the acceleration decreases — the skydiver is still speeding up, but less and less quickly.
3. Eventually the air resistance grows until it exactly balances the weight. Now the resultant force is zero, so by Newton's First Law the skydiver stops accelerating and falls at a constant velocity — the terminal velocity.

The velocity–time graph for a falling skydiver therefore rises steeply at first, then curves as the acceleration decreases, and finally levels off at the terminal velocity.