M2.2 Variable acceleration and vectors with calculus
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Mechanics 2 (A2). Constant-acceleration formulae and constant-velocity vectors are in M1; projectiles are in M2.1. Acceleration given as a function of displacement or velocity (using \(v\,\frac{dv}{dx}\)) is beyond M2.
What this page covers
6 points
- Finding velocity and acceleration by differentiating displacement with respect to time
- Finding velocity and displacement by integrating, using initial conditions to fix the constants
- When a particle is at rest, its greatest (or least) velocity, and the distance travelled when it changes direction
- Polynomial functions of time and powers of \(t\) (including negative and fractional powers), and motion defined in stages
- Position, velocity and acceleration vectors in \(\mathbf{i}\), \(\mathbf{j}\) form: \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\), \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\), and integrating back
- Speed at a given time, moving due north or east, moving parallel to a given vector, and the force from \(\mathbf{F} = m\mathbf{a}\)
In one minute
6 points
- When acceleration changes with time the constant-acceleration formulae stop working, so you differentiate to go from displacement to velocity to acceleration and integrate to come back, doing the same to each component of a vector.
- Differentiate down: \(v = \frac{dx}{dt}\) and \(a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\).
- Integrate up: \(v = \int a\,dt\) and \(x = \int v\,dt\). Every integration needs a constant, fixed by an initial condition.
- At rest: \(v = 0\). Greatest or least velocity: \(\frac{dv}{dt} = 0\), i.e. \(a = 0\), and check the end points of the interval.
- Distance travelled: find when \(v = 0\), split the interval there, and add the sizes of the displacements on each piece.
- Vectors: treat the \(\mathbf{i}\) and \(\mathbf{j}\) components separately. Speed \(= |\mathbf{v}| = \sqrt{p^2 + q^2}\); due north means \(\mathbf{i}\)-component 0 and \(\mathbf{j}\)-component positive; force \(\mathbf{F} = m\mathbf{a}\).
Contents
13 sections
- How do you get velocity and acceleration from displacement?
- How do you work back from acceleration to velocity and displacement?
- How do you find the distance travelled when a particle turns round?
- What changes for negative powers and stage-by-stage motion?
- How do you differentiate a position vector?
- How do you integrate a vector with initial conditions?
- How do you tell when a particle moves parallel to a vector?
- Exam tips for M2.2
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do you get velocity and acceleration from displacement?
A particle P moves along a straight line, and its displacement \(x\) from a fixed point O on the line is a function of time \(t\); positive and negative \(x\) are on opposite sides of O. Velocity is the rate of change of displacement and acceleration is the rate of change of velocity, so you differentiate once for velocity and twice for acceleration.
On a displacement–time graph the velocity is the gradient; on a velocity–time graph the acceleration is the gradient. The particle is instantaneously at rest when \(v = 0\), and if \(v\) changes sign there, it changes direction. The constant-acceleration formulae need \(a\) to be constant: if \(a\) depends on \(t\), they give wrong answers.
A particle P moves along a straight line. At time \(t\) seconds, its displacement from a fixed point O on the line is \(x\) metres, where \(x = 2t^3 - 15t^2 + 24t\), \(t \ge 0\). (a) Find expressions for the velocity and the acceleration of P at time \(t\). (b) Find the times at which P is instantaneously at rest, and its displacement from O at each of these times. (c) Find the acceleration of P at each of these times.
(a) \(v = 6t^2 - 30t + 24\), \(a = 12t - 30\)
(b) \(v = 6(t^2 - 5t + 4) = 6(t - 1)(t - 4) = 0\), so \(t = 1\) or \(t = 4\). Then \(x(1) = 2 - 15 + 24 = 11\) and \(x(4) = 128 - 240 + 96 = -16\). t = 1 s, x = 11 m; t = 4 s, x = −16 m
(c) \(a(1) = 12 - 30\) and \(a(4) = 48 - 30\). −18 m s⁻² and 18 m s⁻²
At \(t = 4\), P is 16 m on the negative side of O: a displacement is a position, not a distance travelled.
How do you work back from acceleration to velocity and displacement?
Integration reverses differentiation: \(v = \int a\,dt\) and \(x = \int v\,dt\). Each indefinite integral has a constant of integration, found from an initial condition such as "starts from rest" (\(v = 0\) when \(t = 0\)) or "passes through O when \(t = 0\)" (\(x = 0\) when \(t = 0\)). The condition may be at a later time, e.g. "\(v = 5\) when \(t = 2\)": substitute and solve for the constant. A definite integral \(\int_{t_1}^{t_2} v\,dt\) gives the change in displacement between two times, the signed area under the \(v\)–\(t\) graph.
For the greatest or least velocity, solve \(\frac{dv}{dt} = 0\), that is \(a = 0\). If \(a\) changes from positive to negative there, \(v\) has a maximum; also check the ends of the time interval. If \(a\) is constant, integrating twice gives \(v = u + at\) and \(s = ut + \tfrac{1}{2}at^2\), so the M1 formulae are a special case.
A particle moves along a straight line. At time \(t\) seconds its acceleration is \((6 - 2t)\) m s⁻². When \(t = 0\) the particle is at the point O on the line and its velocity is 2 m s⁻¹. (a) Find the greatest velocity of the particle. (b) Find its displacement from O at the time when this greatest velocity is reached.
(a) \(v = 6t - t^2 + c\), and \(v = 2\) when \(t = 0\) gives \(c = 2\): \(v = 2 + 6t - t^2\). Greatest velocity where \(a = 0\): \(6 - 2t = 0\), \(t = 3\), and \(v = 2 + 18 - 9 = 11\). It is a maximum: \(a > 0\) before \(t = 3\) and \(a < 0\) after (or \(\frac{d^2v}{dt^2} = -2 < 0\)). 11 m s⁻¹
(b) Integrate again; \(x = 0\) when \(t = 0\) gives a zero constant: \(x = 2t + 3t^2 - \dfrac{t^3}{3}\). At \(t = 3\): \(x = 6 + 27 - 9\). 24 m
Leaving out the constant of integration, or assuming it is zero. "Starts with velocity 2 m s⁻¹" means \(c = 2\) in the velocity, not 0. Write "+ c" every time and then find it.
How do you find the distance travelled when a particle turns round?
\(\int v\,dt\) gives the displacement: motion backwards counts as negative and cancels some of the forward motion. The distance travelled counts every metre, forwards or backwards. So find the times in the interval when \(v = 0\) (and \(v\) changes sign), find the displacement at each of these times and at the ends of the interval, and add the sizes of the changes in displacement on each piece. On the \(v\)–\(t\) graph this is the total area between the curve and the time axis, with areas below the axis counted as positive. A sketch of the motion on a line keeps track of where the particle is and which way it is going.
A particle P starts at O and moves along a straight line. At time \(t\) seconds its velocity is \(v\) m s⁻¹, where \(v = t^2 - 7t + 10\). Find the total distance travelled by P in the interval \(0 \le t \le 6\).
At rest: \((t - 2)(t - 5) = 0\), so \(t = 2\) and \(t = 5\), and \(v\) changes sign at each.
Displacement, with \(x = 0\) when \(t = 0\): \(x = \dfrac{t^3}{3} - \dfrac{7t^2}{2} + 10t\), so \(x(2) = \tfrac{26}{3}\), \(x(5) = \tfrac{25}{6}\) and \(x(6) = 6\).
Distance \(= \tfrac{26}{3} + \left(\tfrac{26}{3} - \tfrac{25}{6}\right) + \left(6 - \tfrac{25}{6}\right) = \tfrac{52}{6} + \tfrac{27}{6} + \tfrac{11}{6}\). 15 m
\(\int_0^6 v\,dt = 6\) m is the displacement, not the distance; it only equals the distance if \(v\) never changes sign.
If a question asks for "the distance travelled", check whether \(v = 0\) inside the interval before you integrate. If it does, show the split explicitly: "P changes direction at \(t = 2\), so distance \(= |x(2) - x(0)| + |x(5) - x(2)| + \ldots\)".
What changes for negative powers and stage-by-stage motion?
Nothing in the method changes, only the powers of \(t\). Use the results from P1 and P2: \(\frac{d}{dt}(t^n) = nt^{n-1}\) for any rational \(n\), e.g. \(\frac{d}{dt}(t^{-2}) = -2t^{-3}\) and \(\frac{d}{dt}(t^{3/2}) = \tfrac{3}{2}t^{1/2}\); and \(\int t^n\,dt = \dfrac{t^{n+1}}{n+1} + c\) for \(n \ne -1\). Rewrite fractions first: \(48/t^2 = 48t^{-2}\) and \(\sqrt{t} = t^{1/2}\). A velocity such as \(12 - 48t^{-2}\) approaches a limiting value (12) but never reaches it, because \(48t^{-2} > 0\) for every \(t\).
Motion is often defined in stages, e.g. one formula for \(0 \le t \le 3\) and another for \(t > 3\). Work each stage with its own formula; the velocity and displacement at the change-over time link the stages. (Exponential or trigonometric functions of time need P3 and P4 calculus and are not in M2.)
A small electric cart moves along a straight track. It is released from rest at O when \(t = 2\), and for \(t \ge 2\) its velocity is \(v\) m s⁻¹, where \(v = 12 - 48t^{-2}\). (a) Find the acceleration of the cart when \(t = 2\). (b) Explain why the speed of the cart never reaches 12 m s⁻¹. (c) Find the time at which the speed is 9 m s⁻¹. (d) Find the distance travelled between \(t = 2\) and \(t = 6\).
(a) \(a = -48 \times (-2)t^{-3} = 96t^{-3}\); at \(t = 2\), \(a = 96 \div 8\). 12 m s⁻²
(b) \(48t^{-2} > 0\) for all \(t\), so \(v < 12\) at every time.
(c) \(12 - 48t^{-2} = 9\) gives \(48t^{-2} = 3\), \(t^2 = 16\). t = 4 s (\(t = -4\) is rejected)
(d) \(v \ge 0\) for \(t \ge 2\) (since \(t^2 \ge 4\)), so distance = displacement: \(\int_2^6 (12 - 48t^{-2})\,dt = \left[12t + 48t^{-1}\right]_2^6 = (72 + 8) - (24 + 24)\). 32 m
A particle P moves along a straight line, starting from rest at O. At time \(t\) seconds its velocity is \(v\) m s⁻¹, where \(v = 4t - t^2\) for \(0 \le t \le 3\) and \(v = \dfrac{27}{t^2}\) for \(t > 3\). (a) Find the greatest speed of P for \(0 \le t \le 3\). (b) Find the distance travelled by P in the first 3 s. (c) Find the acceleration of P when \(t = 6\). (d) Find the total distance travelled by P in the first 6 s.
(a) \(\frac{dv}{dt} = 4 - 2t = 0\) gives \(t = 2\) and \(v = 8 - 4 = 4\); at the ends \(v = 0\) (\(t = 0\)) and 3 (\(t = 3\)). 4 m s⁻¹
(b) \(v = t(4 - t) \ge 0\) on \(0 \le t \le 3\), so there is no change of direction: \(\int_0^3 (4t - t^2)\,dt = \left[2t^2 - \tfrac{t^3}{3}\right]_0^3 = 18 - 9\). 9 m
(c) For \(t > 3\), \(v = 27t^{-2}\), so \(a = -54t^{-3}\); at \(t = 6\), \(a = -54 \div 216\). −0.25 m s⁻² (a deceleration of 0.25 m s⁻²)
(d) For \(t > 3\), \(v > 0\), so again no change of direction: \(\int_3^6 27t^{-2}\,dt = \left[-27t^{-1}\right]_3^6 = -4.5 + 9 = 4.5\). Total \(= 9 + 4.5\). 13.5 m
Both formulae give \(v = 3\) at \(t = 3\), so the velocity is continuous at the change-over: a useful check.
How do you differentiate a position vector?
For motion in a plane, the position vector of P relative to O is \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), with \(x\) and \(y\) functions of \(t\); often \(\mathbf{i}\) points east and \(\mathbf{j}\) north. The unit vectors are constant, so differentiate each component separately: \(\mathbf{v} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}\) and \(\mathbf{a} = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}\).
The direction of motion is the direction of \(\mathbf{v}\), e.g. the angle \(\tan^{-1}(q/p)\) with \(\mathbf{i}\), taking care with the quadrant. Moving due east: \(\mathbf{j}\)-component of \(\mathbf{v}\) zero and \(\mathbf{i}\)-component positive. Due north: \(\mathbf{i}\)-component zero and \(\mathbf{j}\)-component positive. (Due west and south: the other component negative.)
A particle P of mass 0.5 kg moves in a horizontal plane. At time \(t\) seconds its position vector relative to a fixed point O is \(\mathbf{r}\) metres, where \(\mathbf{r} = (2t^2 - 3t)\mathbf{i} + (t^3 - 12t)\mathbf{j}\), \(t \ge 0\). The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are directed due east and due north. (a) Find the speed of P when \(t = 1\). (b) Find the time at which P is moving due east. (c) Find the magnitude of the resultant force on P at that time.
(a) \(\mathbf{v} = (4t - 3)\mathbf{i} + (3t^2 - 12)\mathbf{j}\); at \(t = 1\), \(\mathbf{v} = \mathbf{i} - 9\mathbf{j}\), speed \(= \sqrt{1^2 + 9^2} = \sqrt{82}\). 9.06 m s⁻¹
(b) Due east needs the \(\mathbf{j}\)-component zero: \(3t^2 - 12 = 0\), \(t = 2\) (\(t = -2\) rejected). At \(t = 2\) the \(\mathbf{i}\)-component is \(5 > 0\). t = 2 s
(c) \(\mathbf{a} = 4\mathbf{i} + 6t\mathbf{j}\); at \(t = 2\), \(\mathbf{a} = 4\mathbf{i} + 12\mathbf{j}\) and \(\mathbf{F} = 0.5(4\mathbf{i} + 12\mathbf{j}) = 2\mathbf{i} + 6\mathbf{j}\), so \(|\mathbf{F}| = \sqrt{40}\). 6.32 N
How do you integrate a vector with initial conditions?
Integrate each component separately. The constant of integration is now a constant vector, \(\mathbf{c} = c_1\mathbf{i} + c_2\mathbf{j}\), found from the given initial velocity or position. If you are given the force instead of the acceleration, divide by the mass first: \(\mathbf{a} = \mathbf{F}/m\). The distance of P from O at a given time is \(|\mathbf{r}|\).
A particle of mass 2 kg moves in a horizontal plane under the action of a single force \(\mathbf{F}\) newtons, where \(\mathbf{F} = (12t - 6)\mathbf{i} + 4\mathbf{j}\) at time \(t\) seconds. When \(t = 0\) the particle has velocity \((-6\mathbf{i} + \mathbf{j})\) m s⁻¹ and position vector \((3\mathbf{i} - 2\mathbf{j})\) m relative to a fixed point O. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are directed due east and due north. (a) Find the velocity at time \(t\). (b) Show that the particle is moving due north when \(t = 2\). (c) Find the distance of the particle from O when \(t = 2\).
(a) \(\mathbf{a} = (6t - 3)\mathbf{i} + 2\mathbf{j}\), so \(\mathbf{v} = (3t^2 - 3t)\mathbf{i} + 2t\mathbf{j} + \mathbf{c}\), and \(t = 0\) gives \(\mathbf{c} = -6\mathbf{i} + \mathbf{j}\). \(\mathbf{v} = (3t^2 - 3t - 6)\mathbf{i} + (2t + 1)\mathbf{j}\)
(b) At \(t = 2\): \(\mathbf{v} = (12 - 6 - 6)\mathbf{i} + 5\mathbf{j} = 5\mathbf{j}\): \(\mathbf{i}\)-component zero and \(\mathbf{j}\)-component positive, so due north.
(c) Integrate, with \(\mathbf{r} = 3\mathbf{i} - 2\mathbf{j}\) when \(t = 0\): \(\mathbf{r} = \left(t^3 - \tfrac{3}{2}t^2 - 6t + 3\right)\mathbf{i} + (t^2 + t - 2)\mathbf{j}\). At \(t = 2\), \(\mathbf{r} = -7\mathbf{i} + 4\mathbf{j}\), distance \(= \sqrt{7^2 + 4^2} = \sqrt{65}\). 8.06 m
Using the initial velocity as the constant for the position, or forgetting the second constant. The first constant comes from the initial velocity, the second from the initial position.
How do you tell when a particle moves parallel to a vector?
\(\mathbf{v} = p\mathbf{i} + q\mathbf{j}\) is parallel to \(\alpha\mathbf{i} + \beta\mathbf{j}\) when \(\mathbf{v} = k(\alpha\mathbf{i} + \beta\mathbf{j})\) for some number \(k\), i.e. when \(\beta p = \alpha q\). That gives an equation in \(t\). Solve it, then check the sign of \(k\): if \(k > 0\) the particle moves in the direction of the vector; if \(k < 0\) it moves the opposite way. "Moving parallel to \(\mathbf{j}\)" is the case \(p = 0\). The resultant force at any time is \(\mathbf{F} = m\mathbf{a}\), with magnitude \(\sqrt{F_x^2 + F_y^2}\).
A particle P of mass 0.4 kg moves in a horizontal plane. At time \(t\) seconds its velocity is \(\mathbf{v}\) m s⁻¹, where \(\mathbf{v} = (2t - 3)\mathbf{i} + (t^2 - 6)\mathbf{j}\), \(t \ge 0\). When \(t = 0\), P is at the point with position vector \((2\mathbf{i} + 5\mathbf{j})\) m relative to O. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are directed due east and due north. (a) Find the values of \(t\) for which P is moving parallel to \((\mathbf{i} + 2\mathbf{j})\), and state for which of them P moves in the direction of \(\mathbf{i} + 2\mathbf{j}\). (b) Find the magnitude of the resultant force on P at that time. (c) Find the time at which P is moving due east. (d) Find the distance of P from O when \(t = 3\).
(a) The \(\mathbf{j}\)-component is twice the \(\mathbf{i}\)-component: \(t^2 - 6 = 2(2t - 3)\), so \(t^2 - 4t = 0\), \(t = 0\) or \(t = 4\). At \(t = 0\), \(\mathbf{v} = -3\mathbf{i} - 6\mathbf{j} = -3(\mathbf{i} + 2\mathbf{j})\); at \(t = 4\), \(\mathbf{v} = 5\mathbf{i} + 10\mathbf{j} = 5(\mathbf{i} + 2\mathbf{j})\). t = 0 and t = 4; in the direction of i + 2j at t = 4
(b) \(\mathbf{a} = 2\mathbf{i} + 2t\mathbf{j}\); at \(t = 4\), \(\mathbf{a} = 2\mathbf{i} + 8\mathbf{j}\) and \(\mathbf{F} = 0.8\mathbf{i} + 3.2\mathbf{j}\), so \(|\mathbf{F}| = \sqrt{10.88}\). 3.30 N
(c) Due east: \(t^2 - 6 = 0\) and \(2t - 3 > 0\). \(t = \sqrt{6}\) (negative root rejected), and \(2\sqrt{6} - 3 = 1.90 > 0\). t = √6 ≈ 2.45 s
(d) \(\mathbf{r} = (t^2 - 3t + 2)\mathbf{i} + \left(\tfrac{t^3}{3} - 6t + 5\right)\mathbf{j}\); at \(t = 3\), \(\mathbf{r} = 2\mathbf{i} - 4\mathbf{j}\), distance \(= \sqrt{20} = 2\sqrt{5}\). 4.47 m
A "parallel to" or "due north" equation often gives two roots. Reject negative times, a root where the other component has the wrong sign, and a root where the velocity is zero (a particle at rest is not moving in any direction).
Exam tips for M2.2
Going from \(x \to v \to a\), differentiate. Going from \(a \to v \to x\), integrate and add a constant. Write the chain in the margin before you start.
If the acceleration contains \(t\), \(v = u + at\) and the other M1 formulae do not apply, even when the numbers look reasonable.
\(\frac{dv}{dt} = 0\) finds a turning point of \(v\). The greatest speed may be at an end of the interval, or where \(v\) is most negative. Compare the values at the turning points and at both ends.
Before integrating for a distance, solve \(v = 0\), state whether the particle changes direction in the interval, and split the integral if it does.
"Find the force" wants a vector, e.g. \((2\mathbf{i} + 6\mathbf{j})\) N. "Find the magnitude of the force" or "find the speed" wants a single positive number from Pythagoras.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\int v\,dt\) over an interval always gives the distance travelled. | Backward motion gives negative contributions that cancel forward motion. | Split the interval where \(v = 0\) and add the sizes of the displacements. |
| If \(v = 0\) the acceleration must be zero too. | In worked example 1, \(v = 0\) at \(t = 1\) but \(a = -18\) m s⁻². | A particle can be instantaneously at rest while accelerating; that is how it turns round. |
| The greatest velocity is always where \(a = 0\). | \(a = 0\) gives a stationary value of \(v\), which could be a minimum, and ignores the ends of the interval. | Check the nature of the turning point and compare with the end values. |
| The constant of integration can be left out because it is "just a number". | It carries the initial velocity or position, which changes every later answer. | Include \(+c\) (or \(+\mathbf{c}\) for vectors) and find it from the conditions given. |
| Speed is the sum of the \(\mathbf{i}\) and \(\mathbf{j}\) components of \(\mathbf{v}\). | The components are perpendicular, so they combine by Pythagoras. | Speed \(= \sqrt{p^2 + q^2}\). |
| Any root of "\(\mathbf{v}\) parallel to \(\alpha\mathbf{i} + \beta\mathbf{j}\)" means motion in that direction. | The multiple \(k\) can be negative, giving motion the opposite way. | Substitute each root and check the sign of \(k\). |
How this topic connects
Pure Mathematics P1–P2
Differentiating and integrating powers of \(t\), including negative and fractional powers, are exactly the tools used here; finding maxima with \(\frac{dv}{dt} = 0\) is the stationary-point method.
M1.2 Kinematics and M1.1 Vectors
Constant acceleration is the special case where integrating \(a\) gives the familiar formulae; \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) is what you get by integrating a constant velocity vector.
M2.1 Projectiles and M2.4 Work, energy and power
Integrating \(\mathbf{a} = -9.8\mathbf{j}\) twice gives the vector form of projectile motion; power \(P = Fv\) uses a velocity that may be a function of time.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(v = \frac{dx}{dt}\), \(a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\) | Velocity and acceleration from displacement | No — learn it |
| \(v = \int a\,dt\), \(x = \int v\,dt\) (+ constants) | Velocity and displacement from acceleration, with initial conditions | No — learn it |
| distance \(= \sum |x(t_{k+1}) - x(t_k)|\), split where \(v = 0\) | Distance travelled when the particle changes direction | No — derive it |
| \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\), \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\); \(\mathbf{v} = \int \mathbf{a}\,dt\), \(\mathbf{r} = \int \mathbf{v}\,dt\) | Motion in a plane, component by component | No — learn it |
| speed \(= |\mathbf{v}| = \sqrt{p^2 + q^2}\) | Speed from a velocity vector \(p\mathbf{i} + q\mathbf{j}\) | No — learn it |
| \(p\mathbf{i} + q\mathbf{j} \parallel \alpha\mathbf{i} + \beta\mathbf{j} \iff \beta p = \alpha q\) | Moving parallel to a given vector | No — derive it |
| \(\mathbf{F} = m\mathbf{a}\) | Resultant force from a variable acceleration | No — learn it |
| \(v = u + at\), \(s = ut + \tfrac{1}{2}at^2\), … | Only when \(a\) is constant, not for the variable acceleration on this page | No — must be known |
Check your understanding
1. A particle has displacement \(x = 4t^{3/2}\) metres at time \(t\) seconds, \(t > 0\). Find its velocity and acceleration at time \(t\), and their values when \(t = 4\).
\(v = 6t^{1/2}\) m s⁻¹ and \(a = 3t^{-1/2}\) m s⁻². When \(t = 4\): \(v = 6 \times 2\) and \(a = 3 \div 2\). 12 m s⁻¹ and 1.5 m s⁻²
2. A particle has velocity \(v = 3t^2 - 4t\) m s⁻¹ and is 5 m from O, on the positive side, when \(t = 0\). Find its displacement from O when \(t = 2\).
\(x = t^3 - 2t^2 + c\) with \(c = 5\), so \(x = 8 - 8 + 5\). 5 m (it has moved, but is back where it started).
3. Why can you not use \(s = ut + \tfrac{1}{2}at^2\) for a particle whose acceleration is \(6t\) m s⁻²?
That formula comes from integrating a constant acceleration. Here \(a\) changes with time, so integrate \(6t\) itself: \(v = 3t^2 + c\), and so on.
4. A particle has position vector \(\mathbf{r} = t^2\mathbf{i} + (4t - t^2)\mathbf{j}\). When is it moving parallel to \(\mathbf{i} + \mathbf{j}\), and what is its speed then?
\(\mathbf{v} = 2t\mathbf{i} + (4 - 2t)\mathbf{j}\). Equal components: \(2t = 4 - 2t\), so \(t = 1\) and \(\mathbf{v} = 2\mathbf{i} + 2\mathbf{j}\). t = 1; speed \(2\sqrt{2} \approx 2.83\) m s⁻¹
5. A particle starts at O with velocity \(v = (t - 2)(t - 6)\) m s⁻¹. Find the distance it travels for \(0 \le t \le 8\).
\(x = \tfrac{t^3}{3} - 4t^2 + 12t\). \(x(2) = \tfrac{32}{3}\), \(x(6) = 0\), \(x(8) = \tfrac{32}{3}\). Distance \(= \tfrac{32}{3} + \tfrac{32}{3} + \tfrac{32}{3}\). 32 m, although the final displacement is only \(\tfrac{32}{3}\) m.
Specification checklist
- (M2 1.3) I can differentiate a displacement given as a function of time to find the velocity and the acceleration.
- (M2 1.3) I can integrate an acceleration or a velocity given as a function of time, using initial conditions to find the constants.
- (M2 1.3) I can find when a particle is at rest, and its greatest or least velocity or speed over an interval.
- (M2 1.3) I can find the distance travelled when a particle changes direction, by splitting the interval where \(v = 0\).
- (M2 1.3) I can work with polynomial functions of time and powers of \(t\) (including negative and fractional powers), and with motion defined in stages.
- (M2 1.4) I can differentiate a position vector in \(\mathbf{i}\), \(\mathbf{j}\) form to find velocity and acceleration vectors, and find a speed.
- (M2 1.4) I can integrate acceleration and velocity vectors using an initial velocity and an initial position.
- (M2 1.4) I can find when a particle moves due north, south, east or west, or parallel to a given vector, and find a resultant force using \(\mathbf{F} = m\mathbf{a}\).