Kinematics of Motion in a Straight Line

• State the positive direction first; area below the axis is negative displacement; the formulae need constant acceleration; every integral gets its constant from the start of the motion.
• State the positive direction; gradient of \(s\)–\(t\) is \(v\), gradient of \(v\)–\(t\) is \(a\), area under \(v\)–\(t\) is signed displacement; the formulae only for constant \(a\), the one that leaves out the unwanted quantity; every integral gets its constant from the start.

• 4.2.1 I can understand — understand the concepts of distance and speed as scalar quantities, and of displacement, velocity and acceleration as vector quantities
• 4.2.2 I can sketch — sketch and interpret displacement–time graphs and velocity–time graphs, and in particular appreciate that (a) the area under a velocity–time graph represents displacement (b) the gradient of a displacement–time graph represents velocity (c) the gradient of a velocity–time graph represents acceleration
• 4.2.3 I can use — use differentiation and integration with respect to time to solve simple problems concerning displacement, velocity and acceleration
• 4.2.4 I can use — use appropriate formulae for motion with constant acceleration in a straight line

Accuracy for this chapter. Give non-exact answers to 3 significant figures with units, and keep exact answers exact where the question allows: \(v = \sqrt{444} = 2\sqrt{111}\), \(t = \dfrac{15 \pm 3\sqrt{5}}{2}\). Carry intermediate values unrounded to at least four significant figures (\(21.071\), \(6.0475\), \(4.1459\)) and round once at the end; feeding a rounded \(v = 21.1\) into \(t = \dfrac{v - u}{a}\) gives \(6.07\), not \(6.05\). Show the formula and the substitution before the number: a value from the calculator with no working earns nothing. A speed is never negative; a velocity, displacement or acceleration carries the sign of the direction you stated.

• “The area under the velocity–time graph is always the distance.” Correct Area below the \(t\)-axis is negative displacement: the particle is moving backwards. Displacement = area above − area below; distance = area above + area below. They agree only if the graph never goes below the axis (section B, figure 2).
• “\(a = 6 - 2t\), so \(v = u + at\).” Correct The five formulae need constant acceleration. When \(a\) depends on \(t\), integrate (section C).
• “\(v = \int (6 - 2t)\,dt = 6t - t^2\).” Correct Write \(+\,c\) every time and find it from the initial condition. Here \(c = 0\) only because the particle starts from rest.
• “Distance in the first 4 s = \(s(4) - s(0)\).” Correct That is the displacement. If \(v\) changes sign in the interval, find the rest times and add the size of each change in \(s\) separately.
• “At the highest point, the acceleration is zero.” Correct At the top \(v = 0\); the acceleration is still \(-10\ \text{m s}^{-2}\) (upwards positive), as it is for the whole flight (section D, figure 3).
• “It lands on the ground 20 m below, so \(s = 20\).” Correct With upwards positive, a point below the start has \(s = -20\). Choose a positive direction, write it down, and keep every sign consistent with it.
• “\(a = +2\), so the particle is speeding up.” Correct Compare the signs of \(v\) and \(a\). With \(v = -4\) and \(a = +2\) they differ, so it is slowing down.
• “\(g = 9.8\).” Correct This syllabus uses \(g = 10\ \text{m s}^{-2}\), and so does every calculation in this chapter.
• “It went 5 m forwards and 8 m back, so the distance is 3 m.” Repair Distance is the total path: \(5 + 8 = 13\ \text{m}\). The displacement is \(-3\ \text{m}\).
• “\(a = +2\), so it is speeding up.” Repair Compare the signs of \(v\) and \(a\). \(v = -4\) with \(a = +2\) is slowing down.
• “The area under the graph is \(4 + 9 = 13\), so the displacement is 13 m.” Repair The 9 is below the axis, so it counts as \(-9\): displacement \(4 - 9 = -5\ \text{m}\). 13 m is the distance.
• “The gradient of the \(s\)–\(t\) graph is the acceleration.” Repair The gradient of \(s\)–\(t\) is the velocity; the gradient of \(v\)–\(t\) is the acceleration.
• “\(a = 6 - 2t\) and \(u = 0\), so after 3 s, \(v = u + at = 0\).” Repair The formulae need constant \(a\). Integrate: \(v = 6t - t^2\), so \(v = 9\) at \(t = 3\).
• “\(v = \displaystyle\int (6 - 2t)\,dt = 6t - t^2\).” Repair Write \(+\,c\) and find it from the initial condition. Here \(c = 0\) only because the particle starts from rest; with an initial velocity of 2 it would be \(6t - t^2 + 2\).
• “Distance in the first 4 s \(= s(4) - s(0) = 4\ \text{m}\).” Repair That is the displacement. Solve \(v = 0\) for the rest times, find \(s\) at each, and add the sizes of the changes: 12 m.
• “Greatest velocity where \(v = 0\).” Repair \(v = 0\) is rest. The greatest velocity is where \(a = \dfrac{dv}{dt} = 0\), or at an end of the interval if the question restricts \(t\): compare the end values too.
• “At the top, \(a = 0\).” Repair At the top \(v = 0\); \(a = -10\ \text{m s}^{-2}\) (upwards positive) throughout the flight.
• “It falls to the ground 20 m below, so \(s = 20\)” (with upwards positive). Repair Below the start is \(s = -20\). If you prefer \(s = +20\), make downwards positive and then \(u = -15\) and \(a = +10\): one convention, stated, for the whole problem.
• “\(g = 9.8\).” Repair This syllabus uses \(g = 10\ \text{m s}^{-2}\).
• “\(s = vt - \tfrac{1}{2}at^2\) is in MF19.” Repair MF19 prints four formulae; this fifth one must be known, or \(u\) found from two of the others.
• “\(t^2 - 15t + 45 = 0\) has two roots, so one of them must be wrong.” Repair Both roots are after \(t = 3\), so both are valid: two meetings. Reject a root only with a reason (negative, or before a particle has started).
• “B passes \(O\) 3 s later, so \(s_B = 15t\).” Repair Use one clock. B has been moving from \(O\) for \(t - 3\) seconds, so \(s_B = 15(t - 3)\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 20 (momentum) needs velocity with a sign: re-answer worked example 2. Chapter 21 (Newton's laws) finds an acceleration from the forces and then uses these formulae: re-answer worked example 5 there. Chapter 22 (energy) meets \(v^2 = u^2 + 2as\) again beside the work–energy method: re-answer worked example 6 both ways. Recalling a method inside a new problem is worth more than another pass over this chapter on its own, and the rest of Paper 4 uses this chapter's sign convention throughout.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Mechanics content, examined in Paper 4. Paper 4 (Mechanics) is 40% of an AS Level that includes it and 20% of an A Level that includes it. It assumes knowledge of the algebraic methods of Paper 1, and its calculus (outcome 4.2.3) is restricted to Paper 1 techniques. An A Level route that includes Paper 4 takes Papers 1, 3, 4 and 5; Paper 4 cannot be combined with Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A journey described in words, turned into a velocity–time graph and read for accelerations, distances or an unknown time; an acceleration or velocity given as a function of \(t\), integrated or differentiated with the constant found from the start of the motion, perhaps with a “show that” for a greatest velocity; and constant-acceleration problems in stages, under gravity, or with two particles whose equations are solved together. The reasoning is carried by the positive direction you state and the formula or method you choose.
• MF19's Mechanics entry prints \(v = u + at\), \(s = \tfrac{1}{2}(u + v)t\), \(s = ut + \tfrac{1}{2}at^2\) and \(v^2 = u^2 + 2as\), for uniformly accelerated motion only. \(s = vt - \tfrac{1}{2}at^2\), the value \(g = 10\ \text{m s}^{-2}\), the calculus relationships and the graph facts must all be known. See the MF19 card.
• A sign that contradicts the stated direction (a ground below the start written as \(+20\)); a distance reported where a displacement was asked, or the reverse; a constant of integration assumed to be zero; a rounded velocity fed into the next formula. Keep intermediate values to at least four significant figures, round once to 3 significant figures, keep surds such as \(2\sqrt{111}\) exact where asked, and show the formula and substitution: an unsupported calculator answer earns nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.