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The five constant acceleration formulae (suvat) for motion in a straight line, how to derive them, and their vector form in two dimensions. They are used in almost every mechanics question on Paper 2, often together with forces.
\(v = u + at\) (no \(s\)); \(s = ut + \dfrac{1}{2}at^2\) (no \(v\)); \(s = vt - \dfrac{1}{2}at^2\) (no \(u\)); \(s = \dfrac{1}{2}(u + v)t\) (no \(a\)); \(v^2 = u^2 + 2as\) (no \(t\)).
Method: write down \(s, u, v, a, t\), fill in the three you know, mark the one you want, and use the formula that leaves out the fifth.
Choose a positive direction. \(s\),\(u\), \(v\) and \(a\) are vectors and each takes a sign; \(t\) is positive.
\(s\) is the displacement from the starting point, not the distance travelled. If the object turns round, find when \(v = 0\) and treat each part separately.
A quadratic in \(t\) can have two positive roots: the object is at that displacement twice. Reject negative roots.
Deriving the formulae
diagram
Draw a velocity–time graph: a straight line from \(u\) to \(v\) over time \(t\). The gradient gives \(a = \dfrac{v - u}{t}\), so \(v = u + at\)
The area under it is a trapezium, so \(s = \dfrac{1}{2}(u + v)t\)
Constant acceleration: the gradient gives v = u + at, the area gives s = ½(u + v)t
Substitute \(v = u + at\) into \(s = \dfrac{1}{2}(u + v)t\) to get \(s = ut + \dfrac{1}{2}at^2\); substitute \(u = v - at\) to get \(s = vt - \dfrac{1}{2}at^2\)
Substitute \(t = \dfrac{v - u}{a}\) into \(s = \dfrac{1}{2}(u + v)t\) to get \(2as = v^2 - u^2\), so \(v^2 = u^2 + 2as\)
Vector form
With constant acceleration \(\mathbf{a}\): \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) and \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\), where \(\mathbf{r}_0\) is the starting position.
Collect the \(\mathbf{i}\) and \(\mathbf{j}\) terms; the speed is \(|\mathbf{v}|\). 'Moving parallel to \(\mathbf{i}\)' means the \(\mathbf{j}\) component of \(\mathbf{v}\) is zero.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, A1* = the solution must be fully correct, with every step shown.
Find a quantity using suvat
2 to 4 marksGrade E
Write down \(s, u, v, a, t\) with the values you know, taking the direction of motion as positive.
Choose the formula without the quantity you neither know nor want.
Substitute, solve and give units.
Example. A particle moves in a straight line with constant acceleration. It passes the point \(A\) with speed 4 m s−1 and, 6 s later, passes the point \(B\) with speed 16 m s−1. (a) Find the acceleration of the particle. (b) Find the distance \(AB\)
Use the same origin and the same time \(t\) for both objects.
Write an expression for each displacement in terms of \(t\)
Equate the displacements and solve; \(t = 0\) is just the start.
Substitute back to find the distance.
Example. Car \(A\) passes a point \(O\) with speed 8 m s−1 and moves with constant acceleration 2 m s−2. At the same instant car \(B\) passes \(O\) with constant speed 20 m s−1, moving in the same direction along the same straight road. Find the time taken for \(A\) to catch up with \(B\), and the distance from \(O\) at which this happens.
Write \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) and collect the \(\mathbf{i}\) and \(\mathbf{j}\) components.
Turn conditions into equations: 'parallel to \(\mathbf{i}\)' means the \(\mathbf{j}\) component is zero.
Use \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) for the displacement, adding the starting position if it is not \(O\)
Find distances and speeds as magnitudes.
Example. A particle \(P\) moves with constant acceleration \((2\mathbf{i} - 3\mathbf{j})\) m s−2. At time \(t = 0\), \(P\) is at the origin \(O\) and has velocity \((-4\mathbf{i} + 5\mathbf{j})\) m s−1. (a) Find the velocity of \(P\) when \(t = 3\). (b) Find the value of \(t\) when \(P\) is moving parallel to \(\mathbf{i}\). (c) Find the distance \(OP\) when \(t = 4\)
(c) \(\mathbf{r} = 4(-4\mathbf{i} + 5\mathbf{j}) + \dfrac{1}{2}(16)(2\mathbf{i} - 3\mathbf{j})\) (M1)\(= -4\mathbf{j}\) (A1), so \(OP = 4\) m
(A1)
Don’t lose marks
Using suvat when the acceleration is not constant (for example \(a = 3t\)): use calculus instead.
Forgetting that a deceleration makes \(a\) negative when the velocity is positive.
Treating \(s\) as the distance travelled when the object changes direction.
Not converting km h−1 to m s−1 first.
Forgetting to take the square root after using \(v^2 = u^2 + 2as\)
Rounding in an early stage and carrying the rounded value into later stages.
More tips
Memory tricks
Missing-letter rule: find the one of \(s, u, v, a, t\) that is neither given nor wanted, and use the formula without it.
'Starts from rest' → \(u = 0\); 'comes to rest' → \(v = 0\); 'greatest height' → \(v = 0\)
Stopping distance from speed \(u\) with deceleration \(d\): \(\dfrac{u^2}{2d}\)
Sense check: times must be positive, and a car's stopping distance should be tens of metres, not kilometres.
Exam technique
Write the suvat list at the start of every calculation: it earns method marks and prevents mix-ups.
In a two-stage problem, label the stages and carry the exact final velocity of stage 1 into stage 2.
In 'show that' derivations include every algebraic step: the final A1* is lost if a line is skipped.
Give answers to 3 significant figures, or to 2 or 3 s.f. if you have used \(g = 9.8\)
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Choose and use the correct suvat formulaList \(s, u, v, a, t\), then pick the formula that links the three you know with the one you want.
Grade D
Handle deceleration and direction with signsChoose a positive direction; a deceleration has the opposite sign to the velocity.
Grade C
Solve two-stage and two-object problemsThe final velocity of one stage is the initial velocity of the next; objects meet when their displacements are equal.
Grade B
Derive the constant acceleration formulaeUse the gradient and area of a straight-line velocity–time graph, then eliminate variables.
Grade B
Interpret both roots of a quadratic in \(t\)Two positive roots mean the object is at that displacement twice, for example on the way out and on the way back.
Grade A
Use suvat in vector formUse \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) and \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(\mathbf{i}\) and \(\mathbf{j}\) components.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A tram moves along a straight horizontal track with constant acceleration. It passes the point \(A\) with speed 6 m s−1 and, 4 seconds later, passes the point \(B\) with speed 8 m s−1.
(a) Find the acceleration of the tram.[2]
(b) Find the distance \(AB\).[2]
Show the answer and mark scheme
(a)Answer: \(0.5\) m s−2
M1 for \(8 = 6 + 4a\)
A1 for 0.5 (m s−2)
Worked solution: \(v = u + at \Rightarrow 8 = 6 + 4a \Rightarrow a = 0.5\text{ m s}^{-2}\)
(b)Answer: \(28\) m
M1 for \(s = \frac{(6 + 8)}{2} \times 4\) or another complete method
A1 for 28 (m)
Worked solution: \(s = \left(\frac{u + v}{2}\right)t = \frac{6 + 8}{2} \times 4 = 28\) m
Question 2Medium7 marks
A particle \(P\) moves along a straight line with constant acceleration. At time \(t = 0\), \(P\) passes through the point \(O\) with velocity 12 m s−1. The acceleration of \(P\) is 3 m s−2 in the direction opposite to its initial velocity. The point \(A\) lies on the line, 10.5 m from \(O\) in the direction of the initial velocity of \(P\).
(a) Find the times at which \(P\) is at \(A\).[4]
(b) Explain why there are two answers to part (a).[1]
(c) Find the velocity of \(P\) at the second time it passes \(A\).[2]
Show the answer and mark scheme
(a)Answer: \(t = 1\) and \(t = 7\)
M1 for \(10.5 = 12t - \tfrac{1}{2} \times 3t^2\)
A1 for a correct 3-term quadratic, e.g. \(1.5t^{2} - 12t + 10.5 = 0\)
(b)Answer: \(P\) passes through \(A\) moving away from \(O\), comes instantaneously to rest beyond \(A\), and then passes through \(A\) again on its way back.
B1 for \(P\) passes \(A\) on the way out, stops, and passes \(A\) again moving back towards \(O\)
Worked solution: The acceleration opposes the motion, so \(P\) slows down, stops at a point beyond \(A\) and then returns, passing \(A\) a second time.
(c)Answer: \(-9\) m s−1
M1 for \(v = 12 - 3 \times 7\)
A1 for \(-9\) (or 9 m s−1 towards \(O\))
Worked solution: \(v = u + at = 12 - 3 \times 7 = -9\text{ m s}^{-1}\), so \(P\) is moving towards \(O\) with speed 9 m s−1.
Question 3Hard9 marks
A student drops a stone from rest at the top of a well and hears the splash 2.5 seconds later. Take \(g = 9.8\) m s−2.
(a) In Model 1, the stone falls freely under gravity and the sound of the splash is heard instantly. Use Model 1 to find the depth of the well.[2]
(b) Model 2 also takes into account the time taken for the sound to travel up the well, at a constant speed of 340 m s−1. Show that, if the stone falls for \(T\) seconds, then \(4.9T^2 = 340(2.5 - T)\).[2]
(c) Hence find the depth of the well predicted by Model 2.[3]
(d) Both models ignore air resistance. State, with a reason, whether including air resistance would make the predicted depth greater or smaller.[2]
Show the answer and mark scheme
(a)Answer: 30.6 m
M1 for \(s = \frac{1}{2} \times 9.8 \times 2.5^2\)
A1 for 30.6 (m) (accept 31)
Worked solution: \(s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(9.8)(2.5)^2 = 30.6\) m.
(b)Answer: The depth is \(4.9T^2\) (falling) and also \(340(2.5 - T)\) (the sound travels for the remaining \(2.5 - T\) seconds).
M1 for the depth \(= \frac{1}{2}(9.8)T^2 = 4.9T^2\) and the sound takes \(2.5 - T\) seconds
A1* for equating the two expressions for the depth
Worked solution: The stone falls \(\frac{1}{2}(9.8)T^2 = 4.9T^2\) metres in \(T\) seconds. The sound then travels the same distance up the well in the remaining \(2.5 - T\) seconds, covering \(340(2.5 - T)\) metres. So \(4.9T^2 = 340(2.5 - T)\).
(c)Answer: 28.6 m
M1 for solving \(4.9T^2 + 340T - 850 = 0\)
A1 for \(T = 2.416\) (rejecting the negative root)
A1 for 28.6 (m) (accept 29)
Worked solution: \(4.9T^2 + 340T - 850 = 0 \Rightarrow T = \dfrac{-340 + \sqrt{340^2 + 4(4.9)(850)}}{9.8} = 2.416\) (the other root is negative). Depth \(= 4.9 \times 2.416^2 = 28.6\) m.
(d)Answer: Smaller: air resistance slows the stone, so in the same time it falls a shorter distance.
B1 for smaller
B1 for a reason: air resistance reduces the stone’s acceleration and speed, so it falls less far in the time available
Worked solution: Smaller. With air resistance the stone accelerates less, so it is moving more slowly and falls a shorter distance in the same time.