A-Level — physicsMedium5 min read

SUVAT Equations of Motion

s = ut + ½at²; v = u + at; v² = u² + 2as; selecting and applying the right equation

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# SUVAT Equations of Motion — A-Level Physics

The SUVAT equations describe motion with constant (uniform) acceleration. They connect five variables — displacement, initial velocity, final velocity, acceleration, and time — and are among the most frequently used equations in A-Level Physics.

1. The Five Variables

Symbol Quantity Unit
ss Displacement m
uu Initial velocity m/s
vv Final velocity m/s
aa Acceleration m/s²
tt Time s

2. The Four SUVAT Equations

\boxed{v = u + at} \tag{1}

\boxed{s = ut + \frac{1}{2}at^2} \tag{2}

\boxed{s = \frac{(u + v)}{2} \times t} \tag{3}

\boxed{v^2 = u^2 + 2as} \tag{4}

Choosing the Right Equation

Each equation links four of the five variables. Choose the equation that contains the three known values and the unknown:

Equation Missing Variable
v=u+atv = u + at ss
s=ut+12at2s = ut + \frac{1}{2}at^2 vv
s=(u+v)2ts = \frac{(u+v)}{2}t aa
v2=u2+2asv^2 = u^2 + 2as tt

3. Deriving the Equations

Starting from the definition of acceleration:

a=(v−u)/ta = (v-u)/t → Rearrange: v=u+atv = u + at (Equation 1)

Average velocity = (u+v)/2(u+v)/2 → s=(u+v)2ts = \frac{(u+v)}{2}t (Equation 3)

Substitute (1) into (3): s=u+(u+at)2t=ut+12at2s = \frac{u + (u+at)}{2}t = ut + \frac{1}{2}at^2 (Equation 2)

From (1): t=(v−u)/at = (v-u)/a. Substitute into (3): s=(u+v)(v−u)2a=v2−u22as = \frac{(u+v)(v-u)}{2a} = \frac{v^2-u^2}{2a} → v2=u2+2asv^2 = u^2 + 2as (Equation 4)

4. Sign Conventions

Choose a positive direction (e.g., upwards or rightwards) and be consistent:

  • Displacement in the positive direction: s>0s > 0
  • Velocity in the positive direction: v>0v > 0 or u>0u > 0
  • Acceleration in the positive direction: a>0a > 0
  • For gravity (taking up as positive): a=−9.81a = -9.81 m/s²

5. Free Fall

In free fall (only gravity acting):

  • a=g=9.81a = g = 9.81 m/s² downwards
  • If taking downwards as positive: a=+9.81a = +9.81
  • If taking upwards as positive: a=−9.81a = -9.81

Worked Example: Basic Calculation

Problem

A car accelerates uniformly from 5 m/s to 25 m/s in 8 seconds. Find (a) the acceleration, (b) the displacement.

(a) a=(v−u)/t=(25−5)/8=2.5a = (v-u)/t = (25-5)/8 = 2.5 m/s²

(b) s=(u+v)2t=(5+25)2×8=120s = \frac{(u+v)}{2}t = \frac{(5+25)}{2} \times 8 = 120 m

Solution

Worked Example: Falling Object

Problem

A stone is dropped from rest from a cliff 80 m high. Find the time to reach the ground and the speed on impact. (Take g=9.81g = 9.81 m/s², down as positive.)

u=0u = 0, s=80s = 80 m, a=9.81a = 9.81 m/s²

Time: s=ut+12at2s = ut + \frac{1}{2}at^2 → 80=0+12(9.81)t280 = 0 + \frac{1}{2}(9.81)t^2 → t2=16.31t^2 = 16.31 → t=4.04t = 4.04 s

Speed: v=u+at=0+9.81×4.04=39.6v = u + at = 0 + 9.81 \times 4.04 = 39.6 m/s

Check: v2=u2+2as=0+2(9.81)(80)=1569.6v^2 = u^2 + 2as = 0 + 2(9.81)(80) = 1569.6 → v=39.6v = 39.6 m/s ✓

Solution

Worked Example: Thrown Upwards

Problem

A ball is thrown vertically upwards at 20 m/s. Taking up as positive (a=−9.81a = -9.81 m/s²), find:

(a) Time to reach maximum height: At max height, v=0v = 0. v=u+atv = u + at → 0=20+(−9.81)t0 = 20 + (-9.81)t → t=20/9.81=2.04t = 20/9.81 = 2.04 s

(b) Maximum height: v2=u2+2asv^2 = u^2 + 2as → 0=400+2(−9.81)s0 = 400 + 2(-9.81)s → s=400/19.62=20.4s = 400/19.62 = 20.4 m

(c) Total time of flight (returns to start, s=0s = 0): s=ut+12at2s = ut + \frac{1}{2}at^2 → 0=20t+12(−9.81)t20 = 20t + \frac{1}{2}(-9.81)t^2 → 0=t(20−4.905t)0 = t(20 - 4.905t) t=0t = 0 (start) or t=20/4.905=4.08t = 20/4.905 = 4.08 s

Solution

Worked Example: Braking

Problem

A car travelling at 30 m/s brakes and stops in a distance of 45 m. Find the deceleration.

u=30u = 30, v=0v = 0, s=45s = 45 v2=u2+2asv^2 = u^2 + 2as → 0=900+2a(45)0 = 900 + 2a(45) → a=−900/90=−10a = -900/90 = -10 m/s²

Deceleration = 10 m/s².

Solution

7. Practice Questions

    1. A cyclist accelerates from rest at 0.8 m/s² for 15 seconds. Calculate (a) the final speed, (b) the distance travelled. (4 marks)
    1. A ball is dropped from a tower 45 m high. Calculate the time to reach the ground. (g=9.81g = 9.81 m/s²) (3 marks)
    1. A car decelerates uniformly from 20 m/s to 8 m/s over 100 m. Calculate the deceleration. (3 marks)
    1. A rocket launches vertically from rest with an acceleration of 25 m/s² for 10 s. The engine then cuts out. Calculate the maximum height reached. (5 marks)

    Answers

    1. (a) v=0+0.8(15)=12v = 0 + 0.8(15) = 12 m/s. (b) s=0+12(0.8)(225)=90s = 0 + \frac{1}{2}(0.8)(225) = 90 m.

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Frequently Asked Questions

Can I use SUVAT when acceleration isn't constant?

No — SUVAT only applies to uniform (constant) acceleration. For variable acceleration, use calculus.

What if I need two equations to solve a problem?

Many problems require two SUVAT equations (e.g., one to find tt, then another to find ss). This is common for multi-stage problems.

Summary

  • SUVAT equations apply to constant acceleration only
  • Four equations, each missing one of the five variables
  • Always define a positive direction and use consistent signs
  • Free fall: a=±9.81a = \pm 9.81 m/s² (sign depends on convention)
  • Multi-stage problems: apply SUVAT separately to each stage
Tags:
#a-level#physics#mechanics#suvat#kinematics#equations-of-motion

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