The Workbook

1. Compare with \(y = mx + c\): slope \(m = 3\), intercept \(c = 2\).

2. \(y = 3 \times 4 + 2 = 14\).

3. \(3x + 2 = 0\) gives \(x = -\dfrac{2}{3}\).

Slope \(= \dfrac{8 - 2}{3 - 1} = 3\). Using the point \((1, 2)\): \(y - 2 = 3(x - 1)\), so \(y = 3x - 1\). Check: \(x = 3\) gives \(y = 8\).

1. Intercept \(6\) m/s. Slope \(= \dfrac{0 - 6}{3 - 0} = -2\) m/s\(^2\). So \(v = 6 - 2t\).

2. \(u = 6\) m/s; \(a = -2\) m/s\(^2\), opposite to the initial velocity.

3. For \(t > 3\) s the line goes below the axis: \(v\) is negative. The body has stopped and now moves back the way it came.

1. \(F = kx\) is \(y = mx\): a straight line through the origin with slope \(k\).

2. \(v = u + at\) is \(y = mx + c\): a straight line of slope \(a\), cutting the \(v\)-axis at \(u\), not through the origin.

3. \(K \propto v^2\): a parabola, not a straight line.

4. Against \(v^2\) the same relation is \(K = \dfrac{m}{2}(v^2)\): a straight line through the origin with slope \(\dfrac{m}{2}\).

1. The line passes through the origin, so the slope is \(\dfrac{10}{2} = 5\) V/A.

2. \(V = IR\) is \(y = mx\) with \(m = R\). The slope is the resistance, \(5\ \Omega\).

3. \(V = 5 \times 3 = 15\) V.

\(s = \dfrac{a}{2}(t^2)\) is a straight line in \(t^2\) with slope \(\dfrac{a}{2}\). So \(\dfrac{a}{2} = 5\) and \(a = 10\) m/s\(^2\).

1. \(x^2 - 4 = 0\) gives \(x = \pm 2\).

2. \(x^2 \ge 0\), so \(y\) is least when \(x = 0\): the lowest point is \((0, -4)\).

3. The coefficient of \(x^2\) is positive, so the parabola opens upward.

\(4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4) + 4 = 4 - (x - 2)^2\).

\((x - 2)^2 \ge 0\), so \(y\) is largest, \(4\), at \(x = 2\): the vertex is \((2, 4)\). \(y = 0\) when \((x - 2)^2 = 4\), so \(x = 0\) or \(x = 4\).

1. \(x - \dfrac{x^2}{20} = x\left(1 - \dfrac{x}{20}\right) = 0\) gives \(x = 0\) (launch) or \(x = 20\) m (landing).

2. A parabola is symmetric about its vertex, which lies midway between the roots: \(x = 10\) m. There \(y = 10 - \dfrac{100}{20} = 5\) m.

3. \(\dfrac{dy}{dx} = 1 - \dfrac{x}{10} = 1\) at \(x = 0\). A slope of \(1\) means \(\tan\theta = 1\), so the launch angle is \(45^\circ\).

1. \(5t(4 - t) = 0\) gives \(t = 0\) and \(t = 4\) s. At \(t = 0\) the ball is thrown from the ground; at \(t = 4\) s it returns to the ground.

2. The vertex is midway between the roots, at \(t = 2\) s: \(h = 40 - 20 = 20\) m, the greatest height.

1. \(x = t^2\): parabola.

2. \(x = 2t\): straight line through the origin.

3. \(x = \dfrac{1}{t}\): rectangular hyperbola (\(xt = 1\)).

4. \(y = x^2 + 1\): parabola, shifted up by \(1\).

1. \(x^2 + y^2 = r^2\) with \(r = 5\): centre \((0, 0)\), radius \(5\).

2. \(3^2 + 4^2 = 25\): yes, \((3, 4)\) lies on the circle.

3. \(4^2 + 4^2 = 32 > 25\): the point is farther than \(5\) from the centre, so it is outside.

\((x - a)^2 + (y - b)^2 = r^2\) has centre \((a, b)\) and radius \(r\). So the centre is \((2, -1)\) and \(r = 3\).

Radius \(3\) about \((0, 3)\): \(x^2 + (y - 3)^2 = 9\). The lowest point of this circle is \(3\) below its centre, at \((0, 0)\). The circle touches the \(x\)-axis at the origin.

1. \(x^2 + y^2 = 4\).

2. \(x = 2\cos 60^\circ = 1\) m and \(y = 2\sin 60^\circ = \sqrt{3}\) m. Check: \(1 + 3 = 4\).

3. Arc \(= r\theta\) with \(\theta\) in radians: \(2 \times \dfrac{\pi}{3} = \dfrac{2\pi}{3} \approx 2.1\) m.

1. \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with \(a^2 = 25\) and \(b^2 = 9\): \(a = 5\), \(b = 3\).

2. \(c^2 = 25 - 9 = 16\), so \(c = 4\).

3. The ellipse cuts the \(x\)-axis at \(x = \pm 5\). From the focus at \(x = 4\): nearest distance \(5 - 4 = 1\), farthest \(5 + 4 = 9\), in thousands of kilometres.

Put \(y = x + 1\) into the circle: \(x^2 + (x + 1)^2 = 13\), so \(2x^2 + 2x - 12 = 0\), that is, \(x^2 + x - 6 = 0\).

\((x + 3)(x - 2) = 0\): \(x = 2\) or \(x = -3\). The points are \((2, 3)\) and \((-3, -2)\). Check: \(4 + 9 = 13\) and \(9 + 4 = 13\).

1. \(A\): \(\dfrac{4 - 0}{2} = 2\) m/s. \(B\): the line is flat, \(0\). \(C\): \(\dfrac{0 - 4}{6 - 4} = -2\) m/s.

2. The body moves forward at \(2\) m/s for \(2\) s and stays at rest for \(2\) s. Then it returns at \(2\) m/s and reaches the start at \(t = 6\) s.

3. Distance \(= 4 + 0 + 4 = 8\) m. The body ends where it began, so the displacement is \(0\).

1. \(T^2 = \dfrac{4\pi^2}{g}L\) is \(y = mx\): \(T^2\) is proportional to \(L\), with no constant term. So the graph is a straight line through the origin.

2. Slope \(= \dfrac{4 - 2}{1 - 0.5} = 4\) s\(^2\)/m.

3. \(\dfrac{4\pi^2}{g} = 4\) gives \(g = \pi^2 \approx 9.9\) m/s\(^2\).

1. \(\dfrac{dx}{dt}\): velocity.

2. \(\dfrac{dv}{dt}\): acceleration.

3. \(p = mv\): the slope is the mass \(m\).

4. \(V = IR\): the slope is the resistance \(R\).

1. \(\int v\,dt\): displacement.

2. \(\int a\,dt\): change in velocity.

3. \(\int F\,dx\): work done.

4. \(\int F\,dt\): impulse, which equals the change in momentum.

1. \(d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5\) m.

2. Midpoint \(= \left(\dfrac{1 + 4}{2}, \dfrac{2 + 6}{2}\right) = (2.5, 4)\) m.

The given slope is \(2\), so the perpendicular slope is \(-\dfrac{1}{2}\). Through \((2, 3)\): \(y - 3 = -\dfrac{1}{2}(x - 2)\), so \(y = -\dfrac{x}{2} + 4\).

1. \(PV = \) constant is \(xy = c\): a rectangular hyperbola. \(P\) falls as \(V\) grows, never reaching zero.

2. \(P = 200\left(\dfrac{1}{V}\right)\) is \(y = mx\): a straight line through the origin with slope \(200\) kPa L.

3. \(P = \dfrac{200}{4} = 50\) kPa.

\(\ln N = \ln 1000 - 0.5t\). Taking the exponential: \(N = 1000e^{-0.5t}\). At \(t = 2\) s: \(N = 1000e^{-1} = \dfrac{1000}{e} \approx 368\).

A straight line on a \(\ln N\) graph means exponential decay; the slope is minus the decay constant.